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Same shape, any size
A STORY
The print and the copy
Priya points at a landscape print pinned to the notice board. Dhruv holds a smaller copy of the
same photograph.
“Same hills, same clouds,” Priya says. “Only the size has changed.”
“But the big one is wider and taller,” Dhruv says. “How is that the same shape?”
“Measure them,” Priya says.
Dhruv lays a ruler along his copy. “$15$ cm wide, $10$ cm tall.” He measures the print on the
board. “$30$ cm wide, $20$ cm tall.”
“Every length is exactly double,” Priya says. “And every corner is still a right angle. Equal
angles, and every length grown by the same number: that is what same shape means.”
“Two figures like that are called similar,” Dhruv says.
“Similar,” Priya says. “Same shape, any size.”
“So if I know one length in my copy,” Dhruv asks, “can I find the matching length on the board
without measuring it?”
You can. And for triangles you will find something stronger: the angles alone are enough.
Take a photograph. Enlarge it to poster size. Every angle in the picture stays exactly the same. Only the size changes.
That is what SIMILARITY means for two triangles. Two triangles are similar when their corresponding angles are equal and their corresponding sides are in the same ratio. Check only one of the two conditions, and you can be fooled. Neither condition alone is enough.
Let us look at one result that carries almost every shortcut in this chapter. A line inside a triangle, drawn parallel to one side, divides the other two sides in exactly the same ratio. This is the Basic Proportionality Theorem. We will build AA, SSS and SAS on it, one after another.
Congruence is this chapter’s other main idea. It turns out to be similarity’s special case: the same shape, with the ratio locked at $1 : 1$.
Two triangles show the same shape at two sizes, with matching angle marks and their bases printed as 4 and 6.
The poster is the photograph doubled, so every length is twice as long and the marked angle has not moved at all.
Check yourself
Two triangles are similar when
their corresponding angles are equal and their corresponding sides are in the same ratio
their areas are equal, whatever their shape
their corresponding sides are equal in length
their corresponding sides are in the same ratio, whatever their angles happen to measure at each vertex
Check your answer
✓ their corresponding angles are equal and their corresponding sides are in the same ratio — (A) Similarity needs both conditions together — equal angles and a fixed side ratio.
their areas are equal, whatever their shape — Equal area says nothing about shape — two triangles can share an area with completely different angles.
their corresponding sides are equal in length — Equal side lengths make triangles congruent, a special case; similarity only needs the sides in the SAME RATIO, not equal.
their corresponding sides are in the same ratio, whatever their angles happen to measure at each vertex — Proportional sides alone are not enough — a very flat triangle and a very tall one can share a side ratio without matching angles.
The Basic Proportionality Theorem is the result the AA, SSS and SAS similarity criteria are built on. What does BPT actually guarantee?
a line parallel to one side of a triangle is always exactly half its length, whatever the triangle’s shape
any two sides of a triangle are always in the same ratio
a triangle’s three angles always sum to $180^\circ$
a line parallel to one side of a triangle divides the other two sides in the same ratio
Check your answer
a line parallel to one side of a triangle is always exactly half its length, whatever the triangle’s shape — BPT is about a RATIO between the divided segments, not about the parallel line’s own length compared to the side.
any two sides of a triangle are always in the same ratio — The ratio only holds because of the parallel line cutting the two sides — without it, two sides of a triangle have no fixed ratio.
a triangle’s three angles always sum to $180^\circ$ — That is the angle sum property, a separate fact used elsewhere in the chapter; BPT is specifically about side ratios made by a parallel line.
✓ a line parallel to one side of a triangle divides the other two sides in the same ratio — (D) BPT fixes a ratio between the two divided segments whenever the cutting line is parallel to the third side.
A photograph is enlarged to poster size without cropping. Every distance in the poster is exactly $4$ times the corresponding distance in the photograph, and every angle looks unchanged. This is an everyday case of
congruent figures, since nothing about the picture has changed
neither similar nor congruent, since the sizes differ
similar figures, at a scale factor of $4 : 1$
similar figures, at a scale factor of $1 : 4$
Check your answer
congruent figures, since nothing about the picture has changed — The picture looks the same only because the SHAPE is preserved; the poster is four times the size, so the two are not congruent.
neither similar nor congruent, since the sizes differ — Similarity is exactly the relationship that survives a size change — only congruence needs the sizes to match.
✓ similar figures, at a scale factor of $4 : 1$ — (C) The shape is preserved while every length scales by the same factor — exactly what similarity describes.
similar figures, at a scale factor of $1 : 4$ — The poster is the larger figure, so its distances over the photograph’s give $4:1$, not the reverse.
You have proved triangles congruent before. Now compare triangles that only look alike. Try each check below. It takes a minute.
The triangle congruence tests (Class 9, Ganita Manjari, page 99). Here: a similarity test matches parts in order, as these tests do. Check: Sides $3, 4, 5$ and $5, 3, 4$ pair up, so SSS applies.
RHS congruence (Class 9, Ganita Manjari, page 104). Here: the hypotenuse and one side fix the whole right triangle. Check: Hypotenuse $10$ and a leg $6$ give the other leg $\sqrt{10^2 - 6^2} = 8$.
Sides in the same ratio (Class 8, Ganita Prakash Part 1, page 161). Here: similar triangles need their sides in this same ratio. Check: $2:4$ and $5:10$ are both $1:2$, in proportion.
Area of a triangle (Class 9, Ganita Manjari, page 154). Here: the ratio theorem’s proof compares areas over a common height. Check: A triangle with base $6$ and height $4$ has area $1/2 \cdot 6 \cdot 4 = 12$.
Angles of a triangle (Class 8, Ganita Prakash Part 1, page 94). Here: match two angles and the third matches, since all three make $180^\circ$. Check: a triangle with angles $70^\circ, 60^\circ, 50^\circ$ sums to $180^\circ$.
If any of these felt new, read the page named before going on.
Similarity is not only for triangles. Take any two polygons with the same number of sides. We call them similar when two conditions both hold. Every pair of corresponding angles is equal. Every pair of corresponding sides is in the same ratio.
Check only one of those conditions, and you can be fooled. A square and a rectangle share four equal corresponding angles, all $90^\circ$. But a $2$ cm square’s sides are not in the same ratio as a $2$ cm by $4$ cm rectangle’s sides. Equal angles, unequal ratios: not similar.
Now run the test the other way, and it still fails. A square and a rhombus can have all four sides in the identical ratio $1 : 1 : 1 : 1$. Yet a rhombus can be squashed to any angle at all, so its own angles need not be $90^\circ$. Equal ratios, unequal angles: still not similar.
Try it yourself: a square and a rhombus have sides in the same ratio. Are they similar? (Answer: not necessarily. Their angles need not match.)
Two shape pairs each pass only one similarity test, equal angles alone or equal side ratios alone.
Check yourself
Two polygons with the same number of sides are similar when
their corresponding angles are equal
equal corresponding angles and proportional corresponding sides
their corresponding sides are in the same ratio, whatever their corresponding angles happen to be
they have the same number of sides
Check your answer
their corresponding angles are equal — For polygons in general, equal angles alone are not enough — a square and a rectangle have equal angles but sides not in the same ratio.
✓ equal corresponding angles and proportional corresponding sides — (B) Both conditions are needed together for polygons in general.
their corresponding sides are in the same ratio, whatever their corresponding angles happen to be — For polygons in general, proportional sides alone are not enough — a square and a rhombus have sides in the same ratio but unequal angles.
they have the same number of sides — Having the same number of sides only makes two polygons comparable at all; it is not what makes them similar.
A student says a square and a rectangle are always similar, because every angle in both is $90^\circ$. What is wrong with this claim?
nothing is wrong — matching all four angles is always enough for polygons
equal angles alone do not guarantee the sides are in the same ratio
a square and a rectangle can never be similar, whatever their sides
the angles should be compared in degrees, not right angles
Check your answer
nothing is wrong — matching all four angles is always enough for polygons — Equal angles is only one of the two conditions similarity needs for a polygon; the side ratio must be checked separately.
✓ equal angles alone do not guarantee the sides are in the same ratio — (B) Angle equality is only one of the two conditions similarity needs for a polygon.
a square and a rectangle can never be similar, whatever their sides — A square IS similar to some rectangles — one whose sides are also in a $1:1$ ratio, which makes it a square too.
the angles should be compared in degrees, not right angles — $90^\circ$ and a right angle are the same thing; the unit is not the issue — the missing side-ratio check is.
A square and a rhombus can have all four sides in the same ratio (indeed, equal), yet a student claims this makes them similar. Why is the claim wrong?
a rhombus does not have four sides, so the comparison does not apply
the side ratio must be checked using diagonals, not sides
their corresponding angles are not equal
equal sides always force equal angles in any quadrilateral
Check your answer
a rhombus does not have four sides, so the comparison does not apply — A rhombus does have four sides, all equal — that is exactly why the side-ratio condition looked satisfied here.
the side ratio must be checked using diagonals, not sides — Similarity compares corresponding sides directly; diagonals play no role in the definition.
✓ their corresponding angles are not equal — (C) Equal sides are not enough on their own — the angle condition can still fail.
equal sides always force equal angles in any quadrilateral — A rhombus is the standing counterexample — all four sides equal, yet the angles can differ from $90^\circ$.
Congruence asks for more than similarity does. Two triangles are congruent when they have exactly the same shape and the same size. Every pair of corresponding sides is equal in length. Every pair of corresponding angles is equal in measure.
Think of congruence as similarity with the scale factor pinned down. A similar pair can sit at any ratio at all: $1 : 2$, $1 : 3$, or any other value the problem hands you. A congruent pair is a similar pair whose ratio happens to be exactly $1 : 1$.
A triangle and its exact copy are congruent. A triangle and a poster-sized enlargement of it are similar but not congruent. The shape survived the enlargement. The size did not.
Ask yourself one question to keep the two words apart. Has the size changed, or only the scale of the picture? (Answer: if the size changed, you can say only similar, never congruent.)
Two triangles, one turned and moved, have all three sides and all three angles marked equal to the other’s.
Check yourself
Two triangles are congruent when
their corresponding angles are equal, whatever their side lengths
every corresponding side and angle is equal
their areas are equal, however their shapes differ
their corresponding sides are in the same ratio, at any scale
Check your answer
their corresponding angles are equal, whatever their side lengths — Equal angles alone make triangles similar, not congruent — congruence additionally needs the sides to match exactly.
✓ every corresponding side and angle is equal — (B) Congruence needs every side and every angle to match exactly.
their areas are equal, however their shapes differ — Two differently shaped triangles can share an area; congruence needs every side and angle to match, not just the area.
their corresponding sides are in the same ratio, at any scale — “At any scale” is exactly the similarity condition; congruence fixes the scale at $1:1$.
Congruence is described as the special case of similarity where the scale factor is $1 : 1$. What does this mean in practice?
every similar pair of triangles is automatically a congruent pair too
every congruent pair of triangles is automatically a similar pair too
similarity and congruence describe the same relationship, worded differently
a scale factor of $1:1$ turns any two triangles into similar ones
Check your answer
every similar pair of triangles is automatically a congruent pair too — The implication runs only one way — congruent implies similar, but similar triangles at scale $2:1$, say, are not congruent.
✓ every congruent pair of triangles is automatically a similar pair too — (B) Congruence is a stricter version of similarity, so it always implies similarity, never the reverse.
similarity and congruence describe the same relationship, worded differently — They are related but distinct — similarity allows any scale factor, congruence requires exactly $1:1$.
a scale factor of $1:1$ turns any two triangles into similar ones — The scale factor of $1:1$ describes triangles already similar and happening to match in size; it does not create similarity out of nothing.
Triangle $A B C$ and triangle $D E F$ are similar with scale factor $2 : 1$ ($D E F$ the larger). Are they congruent?
no
yes, since similar triangles are always congruent
yes, since a whole-number scale factor counts as congruent
it cannot be decided without knowing the angles
Check your answer
✓ no — (A) A $2:1$ scale factor means the sizes differ, which rules out congruence.
yes, since similar triangles are always congruent — Similar triangles share shape at any scale; a $2:1$ scale factor means the sizes differ, so they cannot be congruent.
yes, since a whole-number scale factor counts as congruent — No scale factor other than exactly $1:1$ gives congruence, whole number or not.
it cannot be decided without knowing the angles — The two triangles are already given as similar, so their angles already match — the scale factor alone decides congruence.
Apply the general similarity test to triangles, and one more discipline gets added: CORRESPONDENCE ORDER.
Write triangle $A B C$ is similar to triangle $D E F$. That sentence already pairs the vertices. $A$ pairs with $D$, $B$ with $E$, $C$ with $F$. Every statement that follows reads off that pairing. Angle $A$ equals angle $D$, not angle $E$ or angle $F$. Side $A B$ corresponds to side $D E$, not to $E F$ or $F D$.
Change the order of the letters, and the statement changes meaning. Triangle $A B C$ similar to triangle $E D F$ describes a different correspondence entirely, for the very same two triangles. It can be false where the first statement was true.
Before you compare a single angle or side across two triangles said to be similar, write the correspondence out, vertex by vertex. Try it now: triangle $A B C$ is similar to triangle $D E F$. Which angle equals angle $B$? (Answer: angle $E$, the second-listed vertex on each side.) Guessing which vertex matches which, midway through a problem, is the easiest way to compare the wrong pair of angles.
Two panels join the same triangle to D, E, F in one order and E, D, F in the other, crossing the arrows.
Check yourself
For triangle $A B C$ and triangle $D E F$, writing $\triangle A B C ~ \triangle D E F$ means the correspondence is
$A$ with $D$, $B$ with $E$, $C$ with $F$, in that matched order
any pairing of the two triangles’ vertices, since similarity does not depend on order
$A$ with $F$, $B$ with $E$, $C$ with $D$, the reverse order
only the first letters, $A$ with $D$; the rest can be any order
Check your answer
✓ $A$ with $D$, $B$ with $E$, $C$ with $F$, in that matched order — (A) The order the vertices are written in fixes exactly which parts are claimed to match.
any pairing of the two triangles’ vertices, since similarity does not depend on order — The order fixes exactly which angles and sides are claimed to match — it is not arbitrary.
$A$ with $F$, $B$ with $E$, $C$ with $D$, the reverse order — The statement lists $A, B, C$ against $D, E, F$ in the same left-to-right order — reversing it changes which parts are claimed equal.
only the first letters, $A$ with $D$; the rest can be any order — All three positions are fixed together — the statement fixes $B$ with $E$ and $C$ with $F$ just as firmly as $A$ with $D$.
Why does writing $\triangle A B C ~ \triangle D E F$ with the vertices in that exact order matter?
it fixes which sides and angles the ratios and equalities in the statement refer to
it only affects how the statement is read aloud, not its meaning
it is required only for congruent triangles, not for similar ones
it lets the two triangles be drawn in exactly the same orientation and position on the page
Check your answer
✓ it fixes which sides and angles the ratios and equalities in the statement refer to — (A) Every equation drawn from a similarity statement depends on which vertices are paired.
it only affects how the statement is read aloud, not its meaning — The order sets up every equation drawn from the statement — $\angle A = \angle D$, $(A B)/(D E) = (B C)/(E F)$ — so it changes the meaning, not just the reading.
it is required only for congruent triangles, not for similar ones — Vertex correspondence order matters for both similarity and congruence statements equally.
it lets the two triangles be drawn in exactly the same orientation and position on the page — Two similar triangles can be drawn rotated or flipped from each other; correspondence is about which named vertices match, not how the figure looks on the page.
Given $\triangle A B C ~ \triangle D E F$, a student writes the side ratio as $(A B)/(E F) = (B C)/(F D)$. What is the error?
nothing is wrong — any two sides from the two triangles can be compared this way
the sides are paired against the wrong corresponding vertices
the ratio should use angles instead of sides
$D E F$ should have been written before $A B C$ in the similarity statement
Check your answer
nothing is wrong — any two sides from the two triangles can be compared this way — Only ratios that pair CORRESPONDING sides — $A B$ with $D E$, $B C$ with $E F$, $C A$ with $F D$ — are guaranteed equal by the similarity statement.
✓ the sides are paired against the wrong corresponding vertices — (B) Only ratios pairing correctly corresponding sides are guaranteed equal.
the ratio should use angles instead of sides — Sides are the right quantity to compare here; the fix is matching them to the correct corresponding vertices, not switching to angles.
$D E F$ should have been written before $A B C$ in the similarity statement — The similarity statement $\triangle A B C ~ \triangle D E F$ is fine as given; the error is entirely in how the student paired sides for the ratio.
Two quick tests separate genuine similarity from a resemblance that only looks like it. Let us check equilateral triangles first, then isosceles ones.
Every equilateral triangle is similar to every other one, always, with no further checking needed. Each of its three angles measures exactly $60^\circ$, no matter the triangle’s actual size. So the angle condition is satisfied automatically.
Isosceles triangles get no automatic pass like that. Check two examples yourself: a triangle with angles $50^\circ$, $50^\circ$, $80^\circ$ is isosceles, since it has two equal angles. So is a triangle with angles $40^\circ$, $40^\circ$, $100^\circ$. Yet the two sets of angles disagree, so the two triangles are not similar.
Equilateral happens to force the angles too, which is why it always works. Isosceles only fixes that two angles match each other within one triangle. It says nothing about how that triangle compares to one you have not measured yet.
Two equilateral triangles share identical angle marks, but two isosceles triangles have different angles and are not similar.
A SCHOLAR INDIA REMEMBERS
A square and a rectangle both have four right angles. Are they similar? What is your reason? Take a square with sides $1$ and $1$, and a rectangle with sides $1$ and $2$. The angles match. The sides are not in one ratio. So they are not similar. For four-sided shapes, equal angles alone are not enough.
Gautama · the sarus crane
Check yourself
Are all equilateral triangles similar to each other?
no, not unless their side lengths are also equal
yes, since every angle measures $60^\circ$, at any size
only if they are also congruent
it depends on whether the triangles are drawn the same way up
Check your answer
no, not unless their side lengths are also equal — Similarity never requires the sizes to match — only the angles and the side ratio, and equilateral triangles always have both fixed.
✓ yes, since every angle measures $60^\circ$, at any size — (B) A fixed angle set at any size is exactly what similarity requires.
only if they are also congruent — Similarity is the weaker, scale-free condition; two equilateral triangles of different sizes are similar without being congruent.
it depends on whether the triangles are drawn the same way up — Orientation on the page has no bearing on similarity — only the angles and side ratios matter.
Unlike equilateral triangles, two isosceles triangles are not always similar. Why not?
isosceles triangles never have any equal angles to compare
isosceles triangles can have more than three sides
being isosceles fixes equal sides, not the angle measures
similarity does not apply to any triangle with equal sides
Check your answer
isosceles triangles never have any equal angles to compare — An isosceles triangle does have a pair of equal base angles; the issue is that this pair’s measure is not fixed the way an equilateral triangle’s $60^\circ$ is.
isosceles triangles can have more than three sides — An isosceles triangle is still a triangle, three sides always; the point is only two of the three angle values are constrained.
✓ being isosceles fixes equal sides, not the angle measures — (C) Isosceles only guarantees a pair of equal sides, leaving the angle values free to differ.
similarity does not apply to any triangle with equal sides — Equal sides are no obstacle to similarity — an equilateral triangle has three equal sides and is always similar to another; the difference is what value those equal angles take.
One isosceles triangle has angles $50^\circ, 50^\circ, 80^\circ$. A second has angles $40^\circ, 40^\circ, 100^\circ$. Are the two triangles similar?
yes, since both triangles are isosceles
yes, since both triangles’ angles sum to $180^\circ$
it cannot be decided without knowing the side lengths
no, since their corresponding angles do not match
Check your answer
yes, since both triangles are isosceles — Sharing a property such as being isosceles is not the same as being similar — the actual angle values must agree, and here they do not.
yes, since both triangles’ angles sum to $180^\circ$ — Every triangle’s angles sum to $180^\circ$ — true of all triangles, so it cannot be what decides similarity between these two.
it cannot be decided without knowing the side lengths — For triangles, matching corresponding angles is already enough to decide similarity by the AA criterion — no side lengths are needed, and here the angles do not match.
✓ no, since their corresponding angles do not match — (D) Matching angle values, not the shared isosceles label, is what similarity actually needs.
Take two triangles and compare them side by side. Measure every one of the three sides of the first. Match each length against a side of the second.
If all three matched pairs come out equal, the two triangles are congruent. This is the SSS congruence criterion, Side-Side-Side. A triangle with sides $5$ cm, $6$ cm and $7$ cm is congruent to a second triangle with the same three sides, no matter what angles happen to sit between them. You do not need to measure a single angle.
Hinge three straight lengths together at their ends, in any order. Only one triangle can result. There is no second, differently shaped triangle hiding behind the same three lengths.
Check the correspondence carefully when you use SSS. Side $A B$ must match the side reported equal to it in the second triangle, not merely any side of the same length lying elsewhere in it. Your turn: one triangle has sides $3$ cm, $4$ cm, $5$ cm, and a second has sides $3$ cm, $4$ cm, $5$ cm too. Are they congruent by SSS? (Answer: yes. Every side matches its corresponding side.)
Two triangles, moved but not resized, have all three side pairs tick-marked equal, with no angle marked.
Check yourself
The SSS congruence criterion, proved in Class 9, states that two triangles are congruent when
all three sides of one are in the same ratio as all three corresponding sides of the other
all three sides of one equal all three corresponding sides of the other
two sides and the angle between them match
any three matching parts, sides or angles, are equal
Check your answer
all three sides of one are in the same ratio as all three corresponding sides of the other — A matched ratio of sides gives similarity; SSS congruence needs the sides themselves to be EQUAL, not merely proportional.
✓ all three sides of one equal all three corresponding sides of the other — (B) SSS congruence fixes both shape and size through three matched, equal side lengths.
two sides and the angle between them match — Matching two sides and the included angle is the SAS criterion, a different route to the same conclusion.
any three matching parts, sides or angles, are equal — SSS names its three matched parts specifically as sides; three matching angles alone only gives similarity, never congruence.
Why is matching all three sides always enough to make two triangles congruent?
three sides are simply easier to measure in a classroom than three angles would be
three sides guarantee the triangle is equilateral
three sides fix the triangle’s angles but not its size
a triangle’s three side lengths fix its shape and size completely
Check your answer
three sides are simply easier to measure in a classroom than three angles would be — Ease of measurement has nothing to do with why the criterion works — three side lengths leave no freedom left in how the triangle can be drawn.
three sides guarantee the triangle is equilateral — SSS applies to any triangle, equilateral or not — it only requires the three sides of one to match the three corresponding sides of the other, whatever those lengths are.
three sides fix the triangle’s angles but not its size — Three matched side lengths fix both the shape and the size together — that is exactly why SSS gives congruence, not just similarity.
✓ a triangle’s three side lengths fix its shape and size completely — (D) There is no freedom left in how a triangle can be drawn once all three sides are fixed.
Triangle $A B C$ has sides $3, 4, 5$; triangle $D E F$ has sides $6, 8, 10$ — every side of $D E F$ is exactly twice the corresponding side of $A B C$. A student concludes the two triangles are congruent by SSS. What is wrong?
nothing is wrong — matching all three sides, in any ratio, is what SSS means
the triangles cannot be compared at all, since their sides differ in length
the student should have checked the angles instead of the sides
the sides are in the same ratio
Check your answer
nothing is wrong — matching all three sides, in any ratio, is what SSS means — SSS congruence needs the sides to be EQUAL, not merely in a fixed ratio; a ratio of $2:1$ is the SSS similarity criterion instead.
the triangles cannot be compared at all, since their sides differ in length — The triangles ARE related — similar, at scale factor $2:1$ — the error is only in calling that relationship congruence.
the student should have checked the angles instead of the sides — The sides were the right thing to check; the error is in what conclusion a matched ratio of sides supports — similarity, not congruence.
✓ the sides are in the same ratio — (D) A matched ratio of sides is the SSS similarity condition, not the congruence one.
SAS congruence needs three matched parts too, in a specific arrangement. Take two sides of one triangle and the angle sandwiched between them, the INCLUDED angle. Match all three against a second triangle’s corresponding two sides and included angle.
Picture a $5$ cm side, a $60^\circ$ included angle and a $7$ cm side in one triangle. Match them, in that order, against the same three measurements in another triangle. The two triangles are then congruent by SAS.
That single word, INCLUDED, decides whether SAS applies at all. Match two sides and some other angle of the triangle, one not sitting between them, and the criterion does not apply. A different triangle can still share those same three measurements.
Let us picture the construction together. Draw the two sides first, joined at a point. Then fix the angle between them before you draw the sides’ far ends. The order only works because the angle sits between the two sides being drawn. Your turn: does SAS apply to a $5$ cm side, a $7$ cm side and a $60^\circ$ angle that is NOT between them? (Answer: no. The angle must be included.)
The included angle fixes one triangle on the left, but the same two sides with a non-included angle allow two triangles on the right.
Check yourself
The SAS congruence criterion states that two triangles are congruent when
two sides and any one angle of the triangle match two corresponding sides and any one corresponding angle
two sides and the angle included between them match two corresponding sides and their included angle
one side and the two angles adjacent to it match
two angles and any side match two corresponding angles and any side
Check your answer
two sides and any one angle of the triangle match two corresponding sides and any one corresponding angle — SAS specifically needs the matched angle to sit between the two matched sides — any other angle does not fix the triangle the same way.
✓ two sides and the angle included between them match two corresponding sides and their included angle — (B) SAS names its three matched parts in order — side, included angle, side.
one side and the two angles adjacent to it match — Matching a side and its two adjacent angles is the ASA criterion, not SAS.
two angles and any side match two corresponding angles and any side — SAS is named for its parts in order — Side, Angle, Side — with two sides matched, not two angles.
Triangle $A B C$ and triangle $D E F$ have $A B = D E$, $B C = E F$, and angle $A$ equal to angle $D$. A student claims SAS congruence applies. What is wrong?
nothing is wrong — matching any two sides and any one angle of the triangle is always enough
the two sides should have been $A B$ and $A C$ instead
SAS requires all three sides to be checked, not just two
angle $A$ is not the angle included between the two matched sides $A B$ and $B C$
Check your answer
nothing is wrong — matching any two sides and any one angle of the triangle is always enough — SAS needs the included angle specifically — the one between the two matched sides, which is angle $B$ here, not angle $A$.
the two sides should have been $A B$ and $A C$ instead — The sides given, $A B$ and $B C$, are a fine pair to match; the problem is only that the given angle, $A$, is not the one between them.
SAS requires all three sides to be checked, not just two — SAS is built on exactly two sides plus their included angle; requiring a third side would make it SSS, a different criterion.
✓ angle $A$ is not the angle included between the two matched sides $A B$ and $B C$ — (D) SAS requires the matched angle to sit between the two matched sides, and here it does not.
In triangle $P Q R$ and triangle $X Y Z$, sides $P Q = X Y$ and $Q R = Y Z$ are matched. Which angle must also match for SAS to apply?
angle $P$, matched with angle $X$
angle $Q$, matched with angle $Y$
angle $R$, matched with angle $Z$
any one of the three angles, since all give the same result
Check your answer
angle $P$, matched with angle $X$ — Angle $P$ sits at the end of side $P Q$ only, not between both matched sides — it is not the included angle here.
✓ angle $Q$, matched with angle $Y$ — (B) The included angle is the one physically between the two matched sides.
angle $R$, matched with angle $Z$ — Angle $R$ sits at the end of side $Q R$ only; the included angle must touch both matched sides at once, which only angle $Q$ does.
any one of the three angles, since all give the same result — Only the angle physically between the two matched sides fixes the triangle under SAS; the other two angles do not play that role.
ASA congruence swaps the roles SAS uses. Two angles now anchor the match, and a side sits between them. Take two angles of one triangle and the side joining them, the INCLUDED side. Match all three against a second triangle’s corresponding two angles and included side.
Take a triangle with angles $50^\circ$ and $60^\circ$ and a $6$ cm side between them. It is congruent by ASA to any triangle carrying that same $50^\circ$ angle, $6$ cm side and $60^\circ$ angle, in that order.
SAS asks the same discipline of its angle. ASA asks it of a side instead. An angle at each end of the wrong side breaks the criterion, even where every measurement is individually correct. Only the side actually wedged between two matched angles counts.
Once we fix two angles of a triangle, the third is fixed too, by the angle sum property. But check that ASA still names the side explicitly. A side’s own length is never implied by angles alone. Your turn: a triangle has angles $50^\circ$ and $60^\circ$, but the $6$ cm side given is NOT between them. Does ASA apply? (Answer: no. The side must be included.)
Two triangles, moved but not resized, have two matching angles and the side between them marked equal.
Check yourself
The ASA congruence criterion states that two triangles are congruent when
two angles and any side match two corresponding angles and any side
two sides and the angle between them match
two angles and their included side match
all three angles match, with no side needed
Check your answer
two angles and any side match two corresponding angles and any side — ASA specifically needs the matched side to sit between the two matched angles.
two sides and the angle between them match — Matching two sides and their included angle is SAS, not ASA.
✓ two angles and their included side match — (C) ASA names its matched parts in order — angle, included side, angle.
all three angles match, with no side needed — Matching all three angles only gives similarity — ASA additionally needs one matched side to fix the size.
Knowing two angles of a triangle already fixes the third, by the angle sum property — that alone only gives similarity (AA). What does the included side add, to upgrade this to ASA congruence?
it fixes the actual size of the triangle, not just its shape
it fixes a second pair of equal angles, on top of the two already given
it removes the need for the two angles to be equal
it only matters for triangles that are also isosceles
Check your answer
✓ it fixes the actual size of the triangle, not just its shape — (A) The included side is what turns a fixed shape into a fixed size as well.
it fixes a second pair of equal angles, on top of the two already given — The side is a length, not an angle — the two given angles already fix all three angles between them; the side’s job is fixing size, not adding a third angle match.
it removes the need for the two angles to be equal — ASA still needs both angles matched; the side is an additional condition on top of them, not a substitute.
it only matters for triangles that are also isosceles — ASA applies to any triangle, isosceles or not — the included side fixes size for every case, not a special one.
Two triangles satisfy the AA criterion (two matched angles), so they are similar. What extra piece of information, added to the same two angles, would upgrade the pair from similar to congruent?
a third matched angle
one matched side length, specifically the side between the two matched angles
matching the triangles’ areas
nothing more is needed at all; the AA criterion already gives congruence for any pair of triangles
Check your answer
a third matched angle — The third angle is already forced to match once two angles match — adding it gives no new information and cannot fix the size.
✓ one matched side length, specifically the side between the two matched angles — (B) A matched included side is what fixes the scale factor to $1:1$.
matching the triangles’ areas — Area is a consequence of the triangle’s dimensions, not an independent condition — the standard congruence criteria always add a side, not an area.
nothing more is needed at all; the AA criterion already gives congruence for any pair of triangles — AA is a similarity criterion only — it fixes shape at any scale, and an extra matched side is exactly what fixes the scale to $1:1$.
RHS congruence is the one criterion reserved for right triangles alone. The name is a checklist: Right angle, Hypotenuse, Side.
Confirm both triangles carry a right angle first. Then match the HYPOTENUSE, the side opposite the right angle and always the longest side in a right triangle. Match one more side, any one, between the two triangles. Once the right angle is confirmed in both, matching those two lengths is enough to make the triangles congruent.
Take a right triangle with hypotenuse $10$ cm and one leg $6$ cm. It is congruent, by RHS, to any right triangle sharing that same hypotenuse and leg. The third side is then forced to match too. You never need to measure it directly.
Check that first, always, before you reach for it. This is also the one congruence criterion the book returns to directly, in a later chapter’s proof about tangent lengths from a point outside a circle. Your turn: two triangles are not right triangles. Can RHS ever apply to them? (Answer: no. RHS needs a right angle in both triangles first.)
Two right triangles have matching right angles, hypotenuses and one leg marked, with the remaining leg left unmarked.
Check yourself
The RHS congruence criterion states that in two right triangles, matching the hypotenuse and
one other angle makes the two triangles congruent
the two acute angles makes the two triangles congruent
the triangle’s area makes the two triangles congruent
one other side makes the two triangles congruent
Check your answer
one other angle makes the two triangles congruent — RHS is named for its parts — Right angle, Hypotenuse, Side — the third matched part is a side, not an angle.
the two acute angles makes the two triangles congruent — RHS needs only the hypotenuse and one other side, both lengths — no separate angle check is required beyond the given right angle.
the triangle’s area makes the two triangles congruent — Two right triangles can share an area without matching any specific side; RHS is about one named side, not area.
✓ one other side makes the two triangles congruent — (D) RHS is named for Right angle, Hypotenuse, and one other matched Side.
RHS applies only to right triangles. Why can it not be used for a triangle with no right angle?
without a right angle, the hypotenuse and one side no longer fix the third side by the Pythagorean relation
a triangle with no right angle does not have a longest side to call the hypotenuse
RHS is really just SSS in disguise, so it works for any triangle
it can be used for any triangle at all, the word “right” in RHS is just a naming convention, nothing more than that
Check your answer
✓ without a right angle, the hypotenuse and one side no longer fix the third side by the Pythagorean relation — (A) Pythagoras is what turns two known sides into a fixed third side, and it needs a right angle to apply.
a triangle with no right angle does not have a longest side to call the hypotenuse — Every triangle has a longest side; the term hypotenuse specifically names the side opposite a right angle, which is why RHS needs that angle to exist.
RHS is really just SSS in disguise, so it works for any triangle — RHS reduces to SSS only because the right angle lets Pythagoras fix the third side — without the right angle, that third side is not determined at all.
it can be used for any triangle at all, the word “right” in RHS is just a naming convention, nothing more than that — “Right” names an actual required condition — a genuine $90^\circ$ angle — not just a naming convention.
A student applies RHS to two triangles that share a hypotenuse-length and one other matching side, but neither triangle has a right angle. What is wrong with this application?
RHS requires a right angle in both triangles, which is missing here
nothing is wrong — matching the longest side and one other side is always enough
the two triangles should be checked with SAS instead, since RHS never applies to any pair
the matching side should be the shortest side, not just any other side
Check your answer
✓ RHS requires a right angle in both triangles, which is missing here — (A) Without a right angle, there is no hypotenuse to begin with, since the term only applies opposite a right angle.
nothing is wrong — matching the longest side and one other side is always enough — Without a right angle, there is no hypotenuse to begin with — the term itself only applies opposite a right angle.
the two triangles should be checked with SAS instead, since RHS never applies to any pair — RHS is a valid criterion for right triangles specifically; the fix here is confirming a right angle in each triangle, not abandoning RHS altogether.
the matching side should be the shortest side, not just any other side — RHS allows the hypotenuse plus any one other matching side; which specific side is not the issue here — the missing right angle is.
Equal heights are why the two areas match, and that equality is the step every proof of the theorem passes through.
THEOREM (Basic Proportionality Theorem). If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
Picture triangle $A B C$ with a line through point $D$ on $A B$ and point $E$ on $A C$, drawn parallel to $B C$. The theorem says $A D : D B = A E : E C$.
Let us see why. The proof compares the AREAS of triangles built from this configuration. It uses one fact about area: triangles standing on the same base, between the same two parallel lines, always have equal area. Follow the numbered proof below, one justification per step, and check each step against the figure as you go.
Proof
Given. Triangle $A B C$ has a line through $D$ on $A B$ and $E$ on $A C$, parallel to $B C$.
To prove. $A D : D B = A E : E C$.
let $D E$ be drawn parallel to $B C$, with $D$ on $A B$ and $E$ on $A C$ This is the exact configuration the theorem describes. We start by naming it.
join $B E$ and $C D$; draw $D M$ perpendicular to $A C$ and $E N$ perpendicular to $A B$ We add these four lines ourselves. They are not given. We draw them so we can compare triangle areas using perpendicular heights.
the area of triangle $A D E$ equals $1/2 \cdot A D \cdot E N$ This is the standard area formula for a triangle. Here $A D$ is the base and $E N$ is the height.
the area of triangle $B D E$ equals $1/2 \cdot D B \cdot E N$ The same formula again, with the same height $E N$ and base $D B$ this time.
area of triangle $A D E$ : area of triangle $B D E$ = $A D : D B$ In this ratio of the two areas, the common factor $1/2 \cdot E N$ cancels. Only $A D : D B$ is left.
the area of triangle $A D E$ equals $1/2 \cdot A E \cdot D M$, and the area of triangle $D E C$ equals $1/2 \cdot E C \cdot D M$ The same reasoning again, this time using base $A E$ or $E C$ and height $D M$.
area of triangle $A D E$ : area of triangle $D E C$ = $A E : E C$ Here too the common factor $1/2 \cdot D M$ cancels. Only $A E : E C$ is left.
the area of triangle $B D E$ equals the area of triangle $D E C$ *Both triangles stand on the same base $D E$, between the same two parallel lines $D E$ and $B C$.* So their heights from $D E$ are equal, and so their areas are equal too.
$A D : D B = A E : E C$ Triangle $A D E$’s area is common to both ratios above. Its partner triangles are equal in area, by the last step. So the two ratios must be equal. That is what we set out to prove.
■
The left panel shows the full construction, and the right panel isolates just the two triangles sharing the joining segment.
A SCHOLAR INDIA REMEMBERS
Gautama lived in north India around the 2nd century BCE. By tradition, he wrote the Nyaya-sutras. That book sets out an argument as a public procedure. You state your reason. You name a general pattern. You apply the pattern to the case in front of you. Anyone watching can check each step. The proof of the Basic Proportionality Theorem is set out the same way.
Gautama · the sarus crane
Check yourself
In the BPT proof, after drawing $D M$ perpendicular to $A C$ and $E N$ perpendicular to $A B$, what is this construction actually for?
so that triangle $A B C$ itself becomes a right triangle once the lines are drawn
so that $D E$ is confirmed parallel to $B C$
so that $A D$ and $A E$ can be shown equal
so that triangle areas can be compared using a common, matching height
Check your answer
so that triangle $A B C$ itself becomes a right triangle once the lines are drawn — The perpendiculars $D M$ and $E N$ are auxiliary lines added for the proof; they do not alter triangle $A B C$ itself.
so that $D E$ is confirmed parallel to $B C$ — $D E$ parallel to $B C$ is already given at the start of the proof; the perpendiculars are used later, to measure area, not to establish parallelism.
so that $A D$ and $A E$ can be shown equal — The proof never claims $A D$ equals $A E$ — it proves a ratio equality, $A D:D B = A E:E C$, and the perpendiculars set up the areas used to get there.
✓ so that triangle areas can be compared using a common, matching height — (D) The perpendiculars supply the heights the area-ratio argument needs.
The proof states that triangle $B D E$ and triangle $D E C$ have equal area. What makes this true?
they share base $D E$ between the same two parallels
they are both right triangles, so their areas must match
$B D$ and $D C$ happen to be equal in this figure
both triangles share the same three vertices as triangle $A D E$
Check your answer
✓ they share base $D E$ between the same two parallels — (A) Equal base and equal height between two parallels is what forces the equal area.
they are both right triangles, so their areas must match — Nothing in the construction makes $B D E$ or $D E C$ a right triangle; the equal-area step relies only on the shared base and the parallel lines.
$B D$ and $D C$ happen to be equal in this figure — $D$ need not be the midpoint of $A B$; $B D$ and $D C$ are not claimed equal anywhere in the proof.
both triangles share the same three vertices as triangle $A D E$ — $B D E$ and $D E C$ share only the side $D E$, not all three vertices — the equal-area argument rests on that shared base sitting between parallel lines, not on the two triangles being the same.
Suppose $D E$ were NOT parallel to $B C$, but the rest of the construction — $B E$, $C D$, and the perpendiculars — stays the same. Which step of the proof fails first?
the step giving the area of triangle $A D E$ as $1/2 \cdot A D \cdot E N$, using base $A D$
the step giving the ratio of the two areas as $A D : D B$
the final conclusion, $A D : D B = A E : E C$
the step claiming triangle $B D E$ and triangle $D E C$ have equal area
Check your answer
the step giving the area of triangle $A D E$ as $1/2 \cdot A D \cdot E N$, using base $A D$ — That area formula holds for ANY triangle with that base and height — it does not depend on $D E$ being parallel to $B C$ at all.
the step giving the ratio of the two areas as $A D : D B$ — That step only needs the two triangles to share the height $E N$, which the construction gives regardless of whether $D E$ is parallel to $B C$.
the final conclusion, $A D : D B = A E : E C$ — The final conclusion is only wrong as a consequence of an earlier step failing; the argument first breaks at the equal-area claim, which needs the parallel lines directly.
✓ the step claiming triangle $B D E$ and triangle $D E C$ have equal area — (D) That step is the only one that actually depends on $D E$ being parallel to $B C$.
Two earlier steps each give a ratio of areas equal to $A D : D B$ and $A E : E C$ respectively, both built from the area of triangle $A D E$. Which move turns this into the final claim $A D : D B = A E : E C$?
joining $B E$ and $C D$ at the start of the construction
the area formula for triangle $B D E$ alone
substituting the equal areas of triangle $B D E$ and triangle $D E C$ into the two ratios
the given configuration of the parallel line, exactly as stated at the very start of the proof
Check your answer
joining $B E$ and $C D$ at the start of the construction — Joining $B E$ and $C D$ only sets up the triangles the later steps measure; the actual equality of the two ratios comes from the closing substitution.
the area formula for triangle $B D E$ alone — One area formula on its own only gives one area — the ratio equality needs the equal-area fact combined with both earlier ratios.
✓ substituting the equal areas of triangle $B D E$ and triangle $D E C$ into the two ratios — (C) The equal-area fact is what lets the two separate ratios be set equal to each other.
the given configuration of the parallel line, exactly as stated at the very start of the proof — The opening hypothesis only states what is given; the actual proof work — turning two separate ratios into one equality — happens at the closing substitution.
Let us run the Basic Proportionality Theorem in the opposite direction. A new, equally useful statement appears.
The theorem itself starts from a parallel line and concludes a ratio. The CONVERSE starts from a ratio and concludes a parallel line. If a line divides two sides of a triangle in the same ratio, that line is parallel to the third side.
The syllabus states this converse without proof. We use it directly, the same way we will use the AA, SSS and SAS similarity criteria a little further on.
Checking a ratio answers whether a line is parallel. Knowing a line is parallel lets you write down a ratio. Your turn: a line divides two sides of a triangle in the ratio $2 : 3$ on both sides. Is the line parallel to the third side? (Answer: yes, by the converse.)
A line drawn at two unequal ratios, 1.3 and 1.5, tilts visibly away from being parallel to the base.
Check yourself
The converse of the Basic Proportionality Theorem states that
if a line divides two sides of a triangle in the same ratio, the line is parallel to the third side
if a line is parallel to one side of a triangle, it divides the other two sides in the same ratio
if two sides of a triangle are equal, the triangle is isosceles
if a line divides two sides of a triangle in the same ratio, the two triangles formed are congruent
Check your answer
✓ if a line divides two sides of a triangle in the same ratio, the line is parallel to the third side — (A) The converse runs BPT’s logic in the opposite direction, from ratio to parallel line.
if a line is parallel to one side of a triangle, it divides the other two sides in the same ratio — That is the original BPT statement; the converse runs the logic in the other direction, from a ratio to a parallel line.
if two sides of a triangle are equal, the triangle is isosceles — That fact is true but has nothing to do with BPT — the converse of BPT is specifically about a ratio implying a parallel line.
if a line divides two sides of a triangle in the same ratio, the two triangles formed are congruent — The converse’s conclusion is that the line is parallel to the third side — it says nothing about the two smaller triangles being congruent to each other.
In triangle $P Q R$, points $E$ on $P Q$ and $F$ on $P R$ are given. Which two ratios must be compared to decide, by the converse of BPT, whether $E F$ is parallel to $Q R$?
$P E : E Q$ and $P F : F R$
$P E : P Q$ and $P F : P R$
$E Q : F R$ and $P E : P F$
$Q R$ and $E F$ directly, as lengths
Check your answer
✓ $P E : E Q$ and $P F : F R$ — (A) The converse compares each side’s own two segments against each other.
$P E : P Q$ and $P F : P R$ — The converse compares the two segments a point creates on each side — $P E$ against $E Q$, not $P E$ against the entire side $P Q$.
$E Q : F R$ and $P E : P F$ — Each ratio in the converse test comes from one side of the triangle, comparing its own two segments — mixing quantities from different sides does not test parallelism.
$Q R$ and $E F$ directly, as lengths — The converse test never compares $Q R$ and $E F$’s lengths directly — it compares how $E$ and $F$ divide their own sides.
The syllabus states the converse of BPT without a separate proof. Why is that reasonable, given the original theorem is already proved?
the converse is a completely different and unrelated geometric fact that merely happens to share its name
the converse is always automatically true whenever the original theorem is true
the converse only applies to right triangles, so it needs less proof
the converse is the same relationship read in the opposite direction, from ratio to parallel line
Check your answer
the converse is a completely different and unrelated geometric fact that merely happens to share its name — A converse is built from the same statement, with hypothesis and conclusion swapped — it is closely related, not unrelated.
the converse is always automatically true whenever the original theorem is true — A converse can fail even when the original theorem holds — BPT’s converse happens to hold too, but that is a fact about this specific theorem, not an automatic rule.
the converse only applies to right triangles, so it needs less proof — The converse of BPT applies to any triangle, not only right triangles.
✓ the converse is the same relationship read in the opposite direction, from ratio to parallel line — (D) A converse is built from the same statement with hypothesis and conclusion swapped.
In triangle $X Y Z$, a student computes $X E : E Y = 2.1 : 3$ and $X F : F Z = 2 : 3$, sees the two ratios look close, and concludes $E F$ is parallel to $Y Z$. What is wrong?
nothing is wrong — ratios that are close enough always count as equal for this test
the student should have compared $X Y$ and $X Z$ instead of the segment ratios
the converse of BPT cannot be tested numerically at all
the two ratios are not exactly equal, so the converse of BPT does not apply
Check your answer
nothing is wrong — ratios that are close enough always count as equal for this test — The converse of BPT needs the ratios to be exactly equal; $2.1:3$ and $2:3$ are close but not the same, so the line is not guaranteed parallel.
the student should have compared $X Y$ and $X Z$ instead of the segment ratios — Comparing the segment ratios $X E:E Y$ and $X F:F Z$ is the correct method; the actual problem is only that the two values came out unequal.
the converse of BPT cannot be tested numerically at all — Numerical ratio comparison is exactly how the converse of BPT is tested in practice — the student’s method was right, but the values must match exactly.
✓ the two ratios are not exactly equal, so the converse of BPT does not apply — (D) The converse needs an exact ratio match, not an approximate one.
Two matching angles turn out to be all that similarity ever needs, for triangles.
The AA criterion, Angle-Angle, states it plainly. If two angles of one triangle equal two angles of another, the two triangles are similar. You do not need to measure a single side.
This is not an accident. Every triangle’s three angles add to $180^\circ$. So once two angles are fixed, the third is fixed too, automatically equal in both triangles. Three matching angles, in turn, force the sides into the same ratio. AA never has to show that step itself. We state it here without proof, exactly as we stated the converse of the Basic Proportionality Theorem.
Checking a third angle anyway wastes a step. The third angle will always match once the first two do. Your turn: two triangles share two matching angles, $50^\circ$ and $70^\circ$. Are they similar? (Answer: yes, by AA. No side needs checking.)
Two triangles of different sizes have two given angle pairs marked, with the third pair marked in a different colour to show it follows automatically.
A SCHOLAR INDIA REMEMBERS
You say two angles are enough. What was the reason again? Walk me through it. The three angles of a triangle add up to $180^\circ$. Say two angles are $50^\circ$ and $60^\circ$ in both triangles. Then the third angle is $70^\circ$ in both. So all three angles match, and the triangles are similar.
Gautama · the sarus crane
Check yourself
The AA similarity criterion states that two triangles are similar when
two angles of one triangle equal two angles of the other
two angles of one triangle equal two angles of the other, and their included side also matches
one angle of one triangle equals one angle of the other
all three angles of one triangle equal all three angles of the other
Check your answer
✓ two angles of one triangle equal two angles of the other — (A) AA needs exactly two matched angles, no more and no less.
two angles of one triangle equal two angles of the other, and their included side also matches — AA needs only two matching angles — adding a matched side would make the criterion ASA, which proves congruence, a stronger claim.
one angle of one triangle equals one angle of the other — A single matched angle leaves the triangle’s shape unfixed; AA specifically needs two matched angles.
all three angles of one triangle equal all three angles of the other — Once two angles match, the third is forced by the angle sum property — requiring it separately adds nothing new.
Why does matching only two angles, not three, already prove two triangles similar?
because two matched angles always determine a triangle’s exact side lengths directly, without any scaling
because most triangles only have two distinct angle values anyway
the third pair of angles is forced to match too, by the angle sum property
because the third angle is irrelevant to a triangle’s shape
Check your answer
because two matched angles always determine a triangle’s exact side lengths directly, without any scaling — Angles fix the triangle’s shape, not a specific length — that is exactly why AA gives similarity (any scale), not congruence (one fixed size).
because most triangles only have two distinct angle values anyway — A scalene triangle has three different angle values; the reason AA works has nothing to do with how many distinct values a triangle happens to have.
✓ the third pair of angles is forced to match too, by the angle sum property — (C) The angle sum property fixes the third angle the instant the first two are known.
because the third angle is irrelevant to a triangle’s shape — The third angle is not irrelevant — it is fully determined once the first two are fixed, by the angle sum property, which is exactly why checking it separately is unnecessary.
In triangle $A B C$ and triangle $D E F$, angle $A$ equals angle $D$, and angle $B$ equals angle $F$ (not angle $E$). Applying AA correctly, which correspondence is established?
$\triangle A B C ~ \triangle D E F$, keeping the vertices in their original written order regardless
the two triangles cannot be similar, since $B$ and $E$ are not equal
angle $C$ must be found before any correspondence can be decided
$\triangle A B C ~ \triangle D F E$, matching $A$ with $D$, $B$ with $F$, $C$ with $E$
Check your answer
$\triangle A B C ~ \triangle D E F$, keeping the vertices in their original written order regardless — The correspondence must follow the angles that are actually equal — $B$ matches $F$ here, not $E$, so the similarity statement must be written $\triangle A B C ~ \triangle D F E$.
the two triangles cannot be similar, since $B$ and $E$ are not equal — $B$ was never claimed equal to $E$ — it is equal to $F$, and matching correctly against $F$ (not $E$) is exactly what makes the two triangles similar.
angle $C$ must be found before any correspondence can be decided — AA only needs two matched angles to fix the correspondence and conclude similarity — angle $C$’s value follows automatically and is not needed first.
✓ $\triangle A B C ~ \triangle D F E$, matching $A$ with $D$, $B$ with $F$, $C$ with $E$ — (D) The correspondence must follow wherever the equal angles actually are, not the default letter order.
A flagpole $6$ m tall casts a shadow $4$ m long at the same moment a nearby tree casts a shadow $10$ m long. Using AA (both shadows made by the sun’s rays at the same angle, both objects vertical), find the tree’s height.
$15$ m, from $6 : 4 = h : 10$
$6.67$ m, from $6 : 10 = h : 4$
$8$ m, from $4 + 6 - 10 + 8$
$10$ m, since the tree’s shadow equals the tree’s height
Check your answer
✓ $15$ m, from $6 : 4 = h : 10$ — (A) The flagpole’s height-to-shadow ratio carries over to the tree in matching order.
$6.67$ m, from $6 : 10 = h : 4$ — The flagpole’s height-to-shadow ratio must match the tree’s height-to-shadow ratio in the same order: $6:4 = h:10$, not $6:10$.
$8$ m, from $4 + 6 - 10 + 8$ — AA similarity gives a ratio relationship between height and shadow, not a sum or difference of the four given numbers.
$10$ m, since the tree’s shadow equals the tree’s height — Shadow length depends on the sun’s angle and the object’s height together — the flagpole’s own numbers show shadow and height are not equal, so the tree’s shadow cannot simply equal its height either.
On the left all three side ratios agree, so the triangles are similar, but on the right a changed angle breaks the third ratio.
SSS similarity asks the opposite question from SSS congruence. It does not ask whether the sides are EQUAL. It asks whether they are in the SAME RATIO.
Take three sides of one triangle in the same ratio as the three corresponding sides of another, and the two triangles are similar. $A B : D E$, $B C : E F$ and $C A : F D$ might all equal $2 : 3$, for example. The syllabus states this without proof.
All three ratios must agree, in correspondence order, not just two of them. Check this yourself: a triangle with sides in ratio $2 : 3$, $2 : 3$ and $2 : 4$ against another is not similar by SSS. Two ratios matching and a third one drifting is not the same thing as all three matching.
Once you confirm SSS similarity, every pair of corresponding angles comes out equal too, with no separate angle check required. Your turn: three side ratios all equal $3 : 4$. Do the corresponding angles need checking separately? (Answer: no. They are equal automatically.)
Check yourself
The SSS similarity criterion states that two triangles are similar when
all three sides of one equal all three corresponding sides of the other
all three sides of one are in the same ratio as all three corresponding sides of the other
two of the three sides are in the same ratio, and the third side’s ratio can be anything at all
the triangles’ perimeters are in the same ratio
Check your answer
all three sides of one equal all three corresponding sides of the other — Equal sides give congruence; SSS similarity only needs the sides to be in the same ratio, not equal.
✓ all three sides of one are in the same ratio as all three corresponding sides of the other — (B) SSS similarity needs a matched ratio across all three side pairs, not equal lengths.
two of the three sides are in the same ratio, and the third side’s ratio can be anything at all — SSS similarity needs all three side ratios to match — two matching ratios alone do not guarantee the third also matches.
the triangles’ perimeters are in the same ratio — Matching perimeters is a consequence of SSS similarity, not the condition itself — the criterion is checked side by side, not on the total.
Triangle $A B C$ has sides $A B = 4$, $B C = 6$, $C A = 9$. Triangle $D E F$ has sides $D E = 8$, $E F = 12$, $F D = 15$. Two of the three ratios, $A B:D E$ and $B C:E F$, both equal $1:2$. Is SSS similarity satisfied?
no, since $C A : F D = 9 : 15$, which is $3:5$, not $1:2$
yes, since two matching ratios out of three is already enough
yes, since the triangles share the same ratio on their longest sides
it cannot be decided without knowing the angles
Check your answer
✓ no, since $C A : F D = 9 : 15$, which is $3:5$, not $1:2$ — (A) All three ratios must agree, and the third one here breaks the pattern.
yes, since two matching ratios out of three is already enough — SSS similarity needs all three side ratios equal — here the third ratio, $9:15$, breaks the pattern the first two suggested.
yes, since the triangles share the same ratio on their longest sides — Checking only the longest sides is not the SSS test — all three side ratios must be compared, and here they are not all equal.
it cannot be decided without knowing the angles — SSS similarity is checked entirely from the three side ratios — no angle information is needed, and the side ratios here already settle it.
Triangle $A B C$ has sides $6, 8, 10$; triangle $D E F$ has sides $9, 12, 15$. A student compares $A B$ with $E F$, $B C$ with $F D$, and $C A$ with $D E$, gets ratios $6:15$, $8:9$, $10:12$, and concludes the triangles are not similar. What went wrong?
nothing went wrong — the triangles genuinely are not similar
the sides should have been added together before comparing
the sides were paired against the wrong vertices
the two triangles have too many different side lengths to compare
Check your answer
nothing went wrong — the triangles genuinely are not similar — Pairing $A B:D E = 6:9 = 2:3$, $B C:E F = 8:12 = 2:3$, $C A:F D = 10:15 = 2:3$ shows all three ratios do match — the triangles are similar once the sides are paired correctly.
the sides should have been added together before comparing — SSS similarity compares each corresponding side’s own ratio, one pair at a time — sides are never added together for this test.
✓ the sides were paired against the wrong vertices — (C) Pairing sides against the wrong vertices hides a ratio that is really there.
the two triangles have too many different side lengths to compare — A scalene triangle (all sides different) is compared exactly the same way as any other — the correspondence, not the variety of lengths, was the actual issue here.
Triangle $A B C$ has sides $A B = 5$, $B C = 7$, $C A = 8$. Triangle $D E F$ is similar to it, $\triangle A B C ~ \triangle D E F$, with $D E = 15$. Find $E F$.
$17$, adding $10$ to $B C$ since $D E$ added $10$ to $A B$
$21$, since the scale factor is $D E : A B = 3$
$7$, since $B C$ stays the same in a similar triangle
$24$, using a scale factor of $15/5$ applied to $C A$ instead of $B C$
Check your answer
$17$, adding $10$ to $B C$ since $D E$ added $10$ to $A B$ — Similar triangles scale by a fixed ratio, not a fixed amount added to each side — the correct scale factor here is $3$, not $+10$.
✓ $21$, since the scale factor is $D E : A B = 3$ — (B) The same scale factor that turns $A B$ into $D E$ turns $B C$ into $E F$.
$7$, since $B C$ stays the same in a similar triangle — Every side scales by the same factor in similar triangles — $B C$ is not a special exception.
$24$, using a scale factor of $15/5$ applied to $C A$ instead of $B C$ — The scale factor $3$ is correct, but it must be applied to $B C = 7$ to find $E F$, not to $C A$.
SAS similarity is SAS congruence’s looser cousin. A ratio stands in for an equality.
If one angle of a triangle equals one angle of another, and the two sides INCLUDING that angle are in the same ratio in both triangles, the two triangles are similar. The syllabus states this without proof, the same way it states AA and SSS similarity.
That is exactly the discipline SAS congruence already demanded. We carry it over unchanged into the similarity version. An angle that does not sit between the two sides being compared does not satisfy SAS similarity, even where the ratio happens to hold.
Compare this to SAS congruence in your own words. Congruence needs the two sides EQUAL and the angle EQUAL. Similarity needs the two sides in the SAME RATIO and the angle EQUAL. Only the side condition loosens. The angle condition never does. Your turn: two triangles have one matching angle and two sides in ratio $3 : 5$ including that angle. Similar by SAS? (Answer: yes.)
Two triangles of different sizes share one marked angle and two matching side ratios, with the third side left unmarked.
Check yourself
The SAS similarity criterion states that two triangles are similar when
one angle of one equals one angle of the other, and the two sides including that angle are in the same ratio
one angle of one equals one angle of the other, and any two of the sides in the triangle are in the same ratio at all
two sides of one equal two sides of the other, with no angle condition
the two triangles’ areas are in the same ratio
Check your answer
✓ one angle of one equals one angle of the other, and the two sides including that angle are in the same ratio — (A) SAS similarity needs the matched angle to be included between the two ratio-matched sides.
one angle of one equals one angle of the other, and any two of the sides in the triangle are in the same ratio at all — SAS similarity specifically needs the two sides that include the matched angle — any other pair of sides does not fix the same relationship.
two sides of one equal two sides of the other, with no angle condition — Two equal sides alone say nothing about ratio or shape without the included angle tying them together.
the two triangles’ areas are in the same ratio — Matching areas can happen for triangles that are not similar at all; SAS similarity is checked from one matched angle and its two including sides, not from area.
Triangle $A B C$ and triangle $D E F$ have $(A B)/(D E) = (B C)/(E F)$, and angle $A$ equal to angle $D$. A student applies SAS similarity directly. What is the problem?
nothing is wrong — matching one angle and any two sides in ratio is always enough
angle $A$ is not the angle included between sides $A B$ and $B C$
the ratio should have used $C A$ instead of $A B$
SAS similarity requires all three sides in ratio, not just two
Check your answer
nothing is wrong — matching one angle and any two sides in ratio is always enough — SAS similarity needs the matched angle to be the one between the two ratio-matched sides — here that angle is $B$, not $A$.
✓ angle $A$ is not the angle included between sides $A B$ and $B C$ — (B) SAS similarity needs the matched angle to sit between the two ratio-matched sides.
the ratio should have used $C A$ instead of $A B$ — $A B$ and $B C$ are a fine pair of sides to use in the ratio; the problem is only that angle $A$ is not the one included between them.
SAS similarity requires all three sides in ratio, not just two — SAS similarity is built on exactly two sides in ratio plus their included angle; needing all three sides would make it SSS similarity instead.
In triangle $P Q R$ and triangle $X Y Z$, $(P Q)/(X Y) = (Q R)/(Y Z)$, and angle $P$ equals angle $X$. Which angle would need to match instead, for SAS similarity to actually apply here?
angle $R$, matched with angle $Z$
angle $P$ is already the correct angle, no change needed
angle $Q$, matched with angle $Y$
any angle works, since the ratio of sides already fixes similarity
Check your answer
angle $R$, matched with angle $Z$ — Angle $R$ touches only $Q R$ among the two matched sides; the included angle must touch both $P Q$ and $Q R$, which is angle $Q$.
angle $P$ is already the correct angle, no change needed — Angle $P$ sits at the end of $P Q$ only, not between the two matched sides — it is not the included angle for this pair.
✓ angle $Q$, matched with angle $Y$ — (C) The included angle is the one shared by both ratio-matched sides.
any angle works, since the ratio of sides already fixes similarity — Two sides in a matching ratio, without their included angle also matching, do not fix the triangle’s shape — a hinge at that vertex could open to many different triangles.
Triangle $A B C$ has $A B = 4$, $A C = 6$, and angle $A = 50^\circ$. Triangle $D E F$ has $D E = 6$, $D F = 9$, and angle $D = 50^\circ$. Are the triangles similar by SAS?
yes
no, since $A B$ does not equal $D E$
no, since three separate measurements were given instead of two
it cannot be decided without also comparing $B C$ and $E F$
Check your answer
✓ yes — (A) Both matched sides carry the same ratio and the included angles agree.
no, since $A B$ does not equal $D E$ — SAS similarity needs the sides in the same ratio, not equal in length — $4:6$ and $6:9$ both simplify to $2:3$, which is exactly the match required.
no, since three separate measurements were given instead of two — SAS similarity needs exactly two sides plus their included angle, which is what was given — the count of numbers is not the issue.
it cannot be decided without also comparing $B C$ and $E F$ — SAS similarity is complete with two matched-ratio sides plus their included angle; checking $B C$ against $E F$ would be needed for SSS, not SAS.
Three equilateral triangles share identical angle marks, but only the first and third also share the same tick count.
Here is the mix-up this chapter exists partly to prevent.
Two triangles with the same three angles look, at a glance, like the same triangle. It is tempting to call them congruent on the strength of that alone. They are not, necessarily. A triangle with angles $60^\circ$, $60^\circ$, $60^\circ$ and sides $4$ cm each is similar to a triangle with the identical three angles and sides $8$ cm each. But the two are not congruent. One is simply the other, doubled in size.
Congruent needs the scale locked at exactly $1 : 1$: same shape AND same size, not merely the same shape at any size that happens to turn up.
Treating similar and congruent as interchangeable words costs marks, specifically on questions that name which one applies. We can catch this trap with one question: has an actual side length been fixed, or only the angles? A question stating equal angles alone is asking about similarity. A question also fixing an actual side length is asking about congruence. Your turn: a question gives two triangles with equal angles only. Similar, congruent, or both? (Answer: similar only. No side length was fixed.)
Two triangles have the same three angles. Are they congruent?
Weaker. A triangle with angles $60^\circ$, $60^\circ$, $60^\circ$ has sides $4$ cm each. A second
triangle has the identical three angles, with sides $8$ cm each. Same angles, so it is tempting
to call them congruent.
But check the sides: $4$ cm against $8$ cm. They do not match, so the two triangles are not
congruent.
Stronger. The two triangles share the same three angles, so they are similar, with sides in ratio $1 : 2$.
Congruent needs more than equal angles: the same shape AND the same size, scale $1 : 1$. Here the
scale is $1 : 2$, not $1 : 1$, so the triangles are similar but not congruent.
A SCHOLAR INDIA REMEMBERS
Tell me the difference in one line each. Similar means the same shape: the angles match, and the sides are in one ratio. Congruent means the same shape and the same size: that ratio is $1$. So every congruent pair is similar. A similar pair is congruent only when the ratio is $1$.
Gautama · the sarus crane
Check yourself
A student sees two triangles with exactly the same three angles and says they must be congruent, since “similar” and “congruent” feel like the same word. What is the real relationship?
the triangles are always congruent whenever their angles all match
the triangles are neither similar nor congruent until the side lengths are checked and compared first
the triangles are congruent only if drawn in the same orientation
the triangles are similar, but not necessarily congruent, unless their sizes also match
Check your answer
the triangles are always congruent whenever their angles all match — Matching angles alone only guarantees similarity, at any scale — congruence additionally needs the scale factor to be exactly $1:1$.
the triangles are neither similar nor congruent until the side lengths are checked and compared first — For triangles, matching all three angles is already enough to establish similarity by AA — no side check is needed for that conclusion, only for upgrading it to congruence.
the triangles are congruent only if drawn in the same orientation — Orientation on the page never decides similarity or congruence — only the angle measures and, for congruence, the actual side lengths.
✓ the triangles are similar, but not necessarily congruent, unless their sizes also match — (D) Matching angles alone establishes similarity, at any scale, never automatically congruence.
Triangle $A B C$ has angles $70^\circ, 60^\circ, 50^\circ$ and sides $3, 4, 5$ cm. Triangle $D E F$ has the SAME three angles but sides $6, 8, 10$ cm. A student calls them congruent because “the angles are identical.” What is the real relationship?
the triangles are congruent, since matching angles is what congruence actually checks
the triangles are similar, at scale factor $2:1$, but not congruent
the triangles are neither similar nor congruent, since their side lengths differ
it cannot be decided, since the triangles’ orientations were not given
Check your answer
the triangles are congruent, since matching angles is what congruence actually checks — Congruence needs matching sides, not just matching angles — here the sides double from one triangle to the other, so the sizes are different.
✓ the triangles are similar, at scale factor $2:1$, but not congruent — (B) Doubled sides mean the two triangles cannot be the same size, even with identical angles.
the triangles are neither similar nor congruent, since their side lengths differ — The sides here are in a fixed ratio, $2:1$ throughout — that is exactly what similarity allows; only congruence is ruled out by the size difference.
it cannot be decided, since the triangles’ orientations were not given — Orientation is irrelevant here — the angle values and the fixed $2:1$ side ratio are enough to settle both similarity (yes) and congruence (no).
Given that two triangles are already known to be similar, what single extra fact would confirm they are also congruent?
that the scale factor is exactly $1:1$
that their three angles are all equal
that their perimeters are both whole numbers
that both triangles are drawn the same way up on the page
Check your answer
✓ that the scale factor is exactly $1:1$ — (A) The scale factor is the one number that separates similarity from congruence.
that their three angles are all equal — Similar triangles already have all three angles equal by definition — repeating that fact confirms nothing new about size.
that their perimeters are both whole numbers — Whether a perimeter happens to be a whole number has no bearing on whether two similar triangles share the same size.
that both triangles are drawn the same way up on the page — Page orientation never affects whether two triangles are congruent — only the actual side lengths matching does.
A sailmaker cuts a small triangular sail as a test pattern, then cuts a full-size sail with the same three angles but every side $5$ times as long. A customer insists the two sails must be “the same,” since they look identical. What is the accurate description?
the two sails are congruent, since “looking identical” is what congruence means
the two sails are similar, at scale factor $5:1$, but not congruent
the two sails are neither similar nor congruent, since fabric use differs
it cannot be decided without knowing the actual side lengths in metres
Check your answer
the two sails are congruent, since “looking identical” is what congruence means — Looking identical in shape is exactly what similarity captures; congruence additionally requires the actual sizes to match, which they do not here.
✓ the two sails are similar, at scale factor $5:1$, but not congruent — (B) The sails share shape but not size, exactly the similar-not-congruent pattern.
the two sails are neither similar nor congruent, since fabric use differs — How much fabric a sail uses follows from its size, but it is not part of the definition of similarity or congruence — the angles and side ratio are.
it cannot be decided without knowing the actual side lengths in metres — The relationship between the sails is fully decided once the scale factor, $5:1$, is known — the specific unit never changes whether they are similar or congruent.
The Basic Proportionality Theorem gives one ratio: $A D : D B = A E : E C$. Both sides compare a part with the other part of the same side.
It is easy to compare a part with the whole side instead, by mistake. Take triangle $A B C$ with $D$ on $A B$, $E$ on $A C$, and $D E$ parallel to $B C$. If you write $A D : D B = A E : A C$, you have mixed a part with a whole. The right side of that ratio should compare $A E$ with $E C$, not $A E$ with the entire side $A C$.
Let us build the habit that prevents this. Look at the divided side first. Name its two pieces. Then look at the other divided side, and name its two pieces the same way. Only then write the ratio, part against part.
Your turn: one side is split into pieces of $2$ cm and $5$ cm. Which is the correct comparison: the two pieces, $2$ and $5$, or one piece against the full $7$ cm side? (Answer: the two pieces, $2$ and $5$.)
First try
With $A D = 3$, $D B = 6$ and $A E = 4$, I wrote $A D / D B = A E / A C$. That gives $3 / 6 = 4 / A C$, so $A C = 8$ and $E C = 4$ cm.
Second look
Both sides of a ratio must be built the same way. The right line is $A D / D B = A E / E C$, giving $3 / 6 = 4 / E C$, so $E C = 8$ cm.
Match part to part on both sides of the ratio, never part to whole.
In a triangle, one side gives 3 cm and 6 cm, the other gives 4 cm and an unknown piece. Find the unknown piece.
Weaker. The careless line writes $A D : D B = A E : A C$, comparing a part on one side with the whole
of the other side.
Substituting, $3 : 6 = 4 : A C$, so $A C = 8$ cm, and $E C = A C - A E = 8 - 4 = 4$ cm.
Check the two ratios this gives: $A D : D B = 1 : 2$ and $A E : E C = 4 : 4 = 1 : 1$. These are
not equal, so something has gone wrong, even though the theorem promised they would match.
Stronger. The right line writes $A D : D B = A E : E C$, part against part on both sides.
Substituting, $3 : 6 = 4 : E C$, so $E C = 8$ cm.
Check the two ratios now: $A D : D B = 1 : 2$ and $A E : E C = 4 : 8 = 1 : 2$. The two ratios
agree, which is what the theorem promised.
$A D = 1.5$ cm, $D B = 3$ cm, $A E = 1$ cm, with $D E$ parallel to $B C$ These are the given measurements. $D$ is on $A B$ and $E$ is on $A C$.
$A D : D B = A E : E C$ The Basic Proportionality Theorem applies here, since $D E$ is parallel to $B C$.
$1.5 : 3 = 1 : E C$ Substitute the given lengths into the ratio.
$1.5 \cdot E C = 3$ Cross-multiply the two ratios.
$E C = 2$ cm Divide both sides by $1.5$.
$A D : D B = 1.5 : 3 = 0.5$ and $A E : E C = 1 : 2 = 0.5$ Check the answer by confirming both ratios come out equal. They do, so $E C = 2$ cm is correct.
A triangle with a line parallel to its base shows both ratios, 1.5 to 3 and 1 to 2, equal at 0.5.
Find the missing piece of a side, using the ratio theorem.
Name the two points Mark $D$ on one side and $E$ on the other, with $D E$ parallel to the third side.
Write the ratio $A D : D B = A E : E C$. Both sides of the ratio compare the same two pieces.
Put in the numbers Substitute the lengths you know into the ratio, leaving the missing piece as the unknown.
Solve for the missing piece Cross-multiply and divide to find the unknown length.
Check yourself
In triangle $A B C$, $D$ lies on $A B$, $E$ lies on $A C$, and $D E$ is parallel to $B C$. By BPT, which equation correctly relates the segments?
$(A D)/(A B) = (A E)/(E C)$
$(A D)/(D B) = (E C)/(A E)$
$(A D)/(D B) = (A E)/(E C)$
$(A D) \cdot (D B) = (A E) \cdot (E C)$
Check your answer
$(A D)/(A B) = (A E)/(E C)$ — BPT relates the two segments a point creates on a side — $A D$ against $D B$, not $A D$ against the whole side $A B$.
$(A D)/(D B) = (E C)/(A E)$ — The two ratios must be written in the same matched order — $A D:D B$ on one side pairs with $A E:E C$ on the other, not its reciprocal.
✓ $(A D)/(D B) = (A E)/(E C)$ — (C) BPT relates each side’s two own segments, matched in the same order on both sides.
$(A D) \cdot (D B) = (A E) \cdot (E C)$ — BPT is a statement about equal ratios, not about the products of the segments being equal.
In triangle $P Q R$, $S$ lies on $P Q$ and $T$ lies on $P R$, with $S T$ parallel to $Q R$. Given $P S = 4$ cm, $S Q = 6$ cm, $P T = 3$ cm, find $T R$.
$2$ cm, from $P S - P T$
$4.5$ cm, from $6 : 4 = 3 : T R$
$4.5$ cm, from $4 : 6 = 3 : T R$
$8$ cm, from $6 + 3 - 4 + 3$
Check your answer
$2$ cm, from $P S - P T$ — BPT gives a ratio relationship between the two sides’ segments, not a subtraction between numbers taken from different sides.
$4.5$ cm, from $6 : 4 = 3 : T R$ — The ratio must keep $P S:S Q$ in the same order as $P T:T R$ — $4:6$, not $6:4$ — the setup here is wrong even though this particular case lands close.
✓ $4.5$ cm, from $4 : 6 = 3 : T R$ — (C) Setting the two segment ratios equal and solving gives the missing length.
$8$ cm, from $6 + 3 - 4 + 3$ — The correct method sets up a single ratio equation, $4:6 = 3:T R$, rather than adding and subtracting the given numbers.
A student solving for $E C$, given $A D = 1.5$ cm, $D B = 3$ cm, $A E = 1$ cm and $D E$ parallel to $B C$, sets up $(A D)/(A B) = (A E)/(E C)$, using $A B = A D + D B = 4.5$ cm. What is the best fix?
replace $A B$ with $D B$ in the ratio, giving $(A D)/(D B) = (A E)/(E C)$
keep the ratio as set up, since $A B$ is a valid length to use
replace $A E$ with $A C$ instead, leaving the left-hand side of the equation completely unchanged
swap the equation to $(D B)/(A D) = (E C)/(A E)$
Check your answer
✓ replace $A B$ with $D B$ in the ratio, giving $(A D)/(D B) = (A E)/(E C)$ — (A) The left-hand ratio needs the segment $D B$, not the whole side $A B$.
keep the ratio as set up, since $A B$ is a valid length to use — BPT specifically relates the two segments a parallel line creates, $A D$ and $D B$ — using the whole side $A B$ changes the relationship and gives a wrong answer.
replace $A E$ with $A C$ instead, leaving the left-hand side of the equation completely unchanged — The error is in the left-hand ratio, comparing $A D$ to the whole side $A B$ instead of to $D B$; changing the right-hand side does not address that.
swap the equation to $(D B)/(A D) = (E C)/(A E)$ — Inverting both sides keeps the same underlying error, using the whole side instead of the correct segment $D B$ — the fix is substituting the right length, not flipping the fractions.
$P E = 3.9$ cm, $E Q = 3$ cm, $P F = 3.6$ cm, $F R = 2.4$ cm These are the given measurements. $E$ is on $P Q$ and $F$ is on $P R$.
$P E : E Q = 3.9 : 3 = 1.3$ Compute the first ratio.
$P F : F R = 3.6 : 2.4 = 1.5$ Compute the second ratio.
$1.3 \neq 1.5$ The two ratios disagree.
$E F$ is not parallel to $Q R$ The converse of the Basic Proportionality Theorem needs both ratios equal. They are not, so the line fails the test.
$P E \cdot F R = 3.9 \cdot 2.4 = 9.36$ and $E Q \cdot P F = 3 \cdot 3.6 = 10.8$ Check the same conclusion a second way, by cross-multiplying instead of dividing. The two products disagree, confirming $E F$ is not parallel to $Q R$.
Check whether a line is parallel to the third side.
Compute the first ratio Divide the first side’s two pieces, in the same order every time.
Compute the second ratio Divide the second side’s two pieces, in that same order.
Compare the two ratios Simplify both fully before you compare them.
Decide parallel or not Equal ratios mean the line is parallel. Unequal ratios mean it is not.
Check yourself
In triangle $P Q R$, $E$ lies on $P Q$ and $F$ lies on $P R$, with $P E = 3.9$ cm, $E Q = 3$ cm, $P F = 3.6$ cm, $F R = 2.4$ cm. Which two values must be compared to test whether $E F$ is parallel to $Q R$?
$P E : P F$ and $E Q : F R$
$P Q : P R$ and $E Q : F R$
$P E : E Q$ and $P F : F R$
$P E + E Q$ and $P F + F R$
Check your answer
$P E : P F$ and $E Q : F R$ — The converse test compares each side’s own two segments against each other — $P E$ with $E Q$, and separately $P F$ with $F R$ — not a value from one side against a value from the other.
$P Q : P R$ and $E Q : F R$ — The test uses each point’s two segments on its own side, not the whole side length compared against a segment from the other.
✓ $P E : E Q$ and $P F : F R$ — (C) Each side’s own two segments are compared against each other.
$P E + E Q$ and $P F + F R$ — The converse of BPT is a ratio test, not a sum — adding the segments back into the whole side length does not test for parallelism.
In triangle $X Y Z$, $M$ lies on $X Y$ with $X M = 5$ cm, $M Y = 4$ cm, and $N$ lies on $X Z$ with $X N = 7.5$ cm, $N Z = 6$ cm. Is $M N$ parallel to $Y Z$?
no, since $X M$ does not equal $X N$
no, since $5$ and $4$ do not divide evenly
it cannot be decided without measuring $Y Z$ directly
yes
Check your answer
no, since $X M$ does not equal $X N$ — The converse of BPT never requires the two sides’ segments to be equal lengths — only their ratios need to match, and here both ratios equal $1.25$.
no, since $5$ and $4$ do not divide evenly — Whether the numbers divide evenly has no bearing on the ratio test — computing $5/4 = 1.25$ and $7.5/6 = 1.25$ is what settles it.
it cannot be decided without measuring $Y Z$ directly — The converse of BPT decides parallelism purely from the two ratios on the other two sides — the length of $Y Z$ itself is never needed.
✓ yes — (D) Both segment ratios agree, which is exactly the converse’s parallel-line condition.
A student compares $X E : E Y = 6 : 4$ and $X F : F Z = 9 : 6$, notices both simplify to $1.5$, and correctly concludes $E F$ is parallel to $Y Z$ — but writes the reasoning as “the four numbers are all multiples of $3$.” What is the best fix to the reasoning, keeping the correct conclusion?
state that $6/4$ and $9/6$ both equal $1.5$
no fix is needed, since being multiples of $3$ is an equally valid reason
change the conclusion to “not parallel,” since the numbers are different
add a check that $X E + E Y$ equals $X F + F Z$
Check your answer
✓ state that $6/4$ and $9/6$ both equal $1.5$ — (A) The ratio equality, not any shared factor, is the real justification.
no fix is needed, since being multiples of $3$ is an equally valid reason — Being multiples of $3$ is incidental — the converse of BPT is triggered by the two ratios being equal, not by any shared factor among the raw numbers.
change the conclusion to “not parallel,” since the numbers are different — $6, 4, 9, 6$ being different numbers is not the issue — their ratios, $6/4$ and $9/6$, are what must be compared, and both equal $1.5$.
add a check that $X E + E Y$ equals $X F + F Z$ — The converse of BPT has no requirement that the two sides’ total lengths match — only that each side’s own division ratio agrees with the other’s.
$B D = 1.2 \cdot 4 = 4.8$ m This is the distance the girl walked in $4$ seconds, at $1.2$ m/s.
angle $E$ is shared; angle $B$ = angle $D$ = $90^\circ$ Both are right angles, formed where the lamp-post and the girl meet the ground.
triangle $A B E$ is similar to triangle $C D E$ Two matching angles prove it, the AA criterion.
$A B : C D = B E : D E$ Corresponding sides of similar triangles are in the same ratio.
$3.6 : 0.9 = (4.8 + x) : x$ Substitute the lamp-post height, the girl’s height, and $B E = B D + D E = 4.8 + x$, with the shadow length $D E = x$.
$3.6 : 0.9$ simplifies to $4 : 1$, so $4.8 + x = 4 x$ Cross-multiply the simplified ratio.
$3 x = 4.8$, so $x = 1.6$ m Solve for $x$.
$(4.8 + 1.6) : 1.6 = 6.4 : 1.6 = 4$ *Check by substituting $x = 1.6$ back into the ratio.* It matches $3.6 : 0.9$, which also simplifies to $4$, so the shadow length is correct.
A 3.6 m lamp-post and a 0.9 m girl share one ray of light, giving similar triangles with matching right angles.
her shadow
the lamp
the lamp-post
Tara walks away from a lit lamp-post at dusk, her shadow falling ahead of her on the pavement.
Use two matching angles to find a missing length.
Find the shared angle Look for one angle the two triangles have in common.
Find the matching right angles Two right angles, one in each triangle, give the second matching angle.
Confirm similarity by AA Two matching angles are enough. The triangles are similar.
Write the ratio of sides Match corresponding sides in the same order as the angles.
Solve for the unknown Cross-multiply the ratio and solve for the missing length.
Check yourself
In the lamp-post and shadow problem, triangle $A B E$ (lamp-post and shadow tip) and triangle $C D E$ (girl and shadow tip) are shown similar by AA. Which two angle facts establish this?
the lamp-post and the girl are the same height
the two shadows are the same length
they share angle $E$, and angle $B$ equals angle $D$, both right angles
the lamp’s angle changes noticeably between the lamp-post triangle and the girl’s triangle
Check your answer
the lamp-post and the girl are the same height — AA is established from angles, not heights — the lamp-post and girl have very different heights, which is exactly why the problem is worth solving.
the two shadows are the same length — The shadow lengths are what the problem is trying to relate, using AA — they are not themselves part of the angle argument that establishes similarity.
✓ they share angle $E$, and angle $B$ equals angle $D$, both right angles — (C) One shared angle plus one matching right angle is exactly what AA needs.
the lamp’s angle changes noticeably between the lamp-post triangle and the girl’s triangle — The whole method depends on the lamp’s light making the same angle at both triangles at that instant — shared angle $E$ is exactly this fact, not a changing one.
A pole $2$ m tall casts a shadow $1.5$ m long. At the same moment, a building casts a shadow $12$ m long. Using AA similarity the same way as the lamp-post problem, find the building’s height.
$9$ m, from $2 : 1.5 = 12 : h$
$13.5$ m, from $2 + 1.5 + 12 - 2$
$16$ m, from $2 : 1.5 = h : 12$
$12$ m, since the building’s height should equal its shadow length
Check your answer
$9$ m, from $2 : 1.5 = 12 : h$ — The building’s height corresponds to the pole’s height in the same position both ratios share — height over shadow on both sides, not height over shadow on one side and shadow over height on the other.
$13.5$ m, from $2 + 1.5 + 12 - 2$ — AA similarity gives a proportional relationship between height and shadow — it does not support adding or subtracting the given lengths directly.
✓ $16$ m, from $2 : 1.5 = h : 12$ — (C) The pole’s height-to-shadow ratio carries over to the building in the same order.
$12$ m, since the building’s height should equal its shadow length — The pole’s own numbers already show height and shadow are not equal ($2$ m tall, $1.5$ m shadow) — there is no reason the building’s would be either.
The lamp-post and shadow argument depends on both the lamp-post and the girl standing exactly vertical, so that angle $B$ and angle $D$ are both right angles. If the girl leaned at a slant instead, which part of the AA argument breaks?
the shared angle $E$ at the shadow tip no longer exists
the two triangles would no longer share any side
the problem would need SAS instead of AA
angle $B$ no longer equals angle $D$
Check your answer
the shared angle $E$ at the shadow tip no longer exists — The angle at the shadow tip, $E$, is formed by the lamp’s light reaching both objects — it is unaffected by whether the girl stands straight or leans.
the two triangles would no longer share any side — AA similarity is about matching angles, not shared sides — the actual break is that the right-angle match at $B$ and $D$ no longer holds.
the problem would need SAS instead of AA — The fix is not a different criterion — it is that one of the two angle facts AA needs, the right angle at $D$, is no longer true once the girl leans.
✓ angle $B$ no longer equals angle $D$ — (D) A slant removes the matching right angle the argument depends on.
Three matching ratios are enough on their own: once the sides say similar, the angles follow without measuring.
Worked example
SSS finds a missing angle
$A B : D E = 4 : 8$, $B C : E F = 6 : 12$, $C A : F D = 5 : 10$ Form the three side ratios, in correspondence order.
all three ratios equal $1 : 2$ Simplify each ratio.
triangle $A B C$ is similar to triangle $D E F$ All three side ratios match, the SSS criterion.
angle $C$ = angle $F$ Corresponding angles of similar triangles are equal.
angle $C = 180^\circ - 80^\circ - 60^\circ = 40^\circ$ The angle sum property, using the given angle $A$ and angle $B$.
angle $F = 40^\circ$ Equal to angle $C$, already found.
angle $D$ + angle $E$ + angle $F$ = $80^\circ + 60^\circ + 40^\circ = 180^\circ$ Check the answer against the angle sum property. The three angles of triangle $D E F$ add to $180^\circ$, confirming angle $F = 40^\circ$ is correct.
Use three side ratios to find a missing angle.
Form the three ratios Match each side of one triangle to its corresponding side in the other, in order.
Simplify each ratio Check that all three come out to the same value.
Confirm similarity by SSS Equal ratios on all three sides prove the triangles are similar.
Read off the angle Use the matching correspondence and the angle sum property to find the missing angle.
Check yourself
Triangle $A B C$ has sides $4, 6, 5$; triangle $D E F$ has sides $8, 12, 10$. What confirms $\triangle A B C ~ \triangle D E F$ by SSS, before angle $F$ is found?
the two triangles have the same perimeter ratio as their longest sides
angle $A$ is given as $80^\circ$
all three ratios, $A B:D E$, $B C:E F$, $C A:F D$, equal $1:2$
$D E F$’s sides are all even numbers
Check your answer
the two triangles have the same perimeter ratio as their longest sides — SSS similarity is confirmed by checking all three side ratios individually, not by a shortcut using only the longest side or the perimeter.
angle $A$ is given as $80^\circ$ — The angle values are used after similarity is established, to find the missing angle — SSS similarity itself is confirmed from the sides alone.
✓ all three ratios, $A B:D E$, $B C:E F$, $C A:F D$, equal $1:2$ — (C) All three side ratios matching is the SSS similarity evidence, ahead of any angle work.
$D E F$’s sides are all even numbers — Whether the side lengths happen to be even numbers has no bearing on SSS similarity — only whether the three ratios match does.
Triangle $P Q R$ has sides $P Q = 3$, $Q R = 4$, $R P = 5$, with angle $Q = 90^\circ$ (the largest angle here, opposite the longest side $R P$). Triangle $X Y Z$ has sides $X Y = 6$, $Y Z = 8$, $Z X = 10$. By SSS similarity, find angle $Y$.
$45^\circ$, half of angle $Q$ since the sides are half as long
it cannot be found without measuring angle $Y$ directly
$60^\circ$, by the angle sum property applied to the other two angles
$90^\circ$
Check your answer
$45^\circ$, half of angle $Q$ since the sides are half as long — Angles do not scale with a similarity ratio — only the sides change by the scale factor; corresponding angles stay exactly equal.
it cannot be found without measuring angle $Y$ directly — Once SSS similarity is confirmed, every corresponding angle is already known — angle $Y$ must equal angle $Q$, $90^\circ$, with no separate measurement needed.
$60^\circ$, by the angle sum property applied to the other two angles — There is no need to derive angle $Y$ from the other two angles of $X Y Z$ — SSS similarity directly hands over angle $Q$’s value, $90^\circ$, to its corresponding angle $Y$.
✓ $90^\circ$ — (D) The correspondence hands angle $Q$’s value straight over to angle $Y$.
Continuing the previous problem (triangle $P Q R$, sides $3,4,5$; triangle $X Y Z$, sides $6,8,10$), a student pairs $P Q$ with $Y Z$ instead of $X Y$, gets the ratio $3:8$, and concludes the triangles are not similar. What is the actual source of the error?
the sides were paired against the wrong vertices
the triangles genuinely are not similar, since $3:8$ does not simplify nicely
the side lengths should be squared before comparing
$3, 4, 5$ and $6, 8, 10$ are unrelated sets of numbers
Check your answer
✓ the sides were paired against the wrong vertices — (A) Correct pairing recovers the same $1:2$ ratio the mismatch hid.
the triangles genuinely are not similar, since $3:8$ does not simplify nicely — Pairing correctly — $P Q:X Y = 3:6$, $Q R:Y Z = 4:8$, $R P:Z X = 5:10$ — gives $1:2$ throughout, confirming similarity; the mismatch, not the triangles, was the problem.
the side lengths should be squared before comparing — SSS similarity compares side lengths directly as ratios — there is no squaring step in the criterion.
$3, 4, 5$ and $6, 8, 10$ are unrelated sets of numbers — $6, 8, 10$ is exactly $2$ times $3, 4, 5$ — the sets are directly related once matched to their correct corresponding sides.
$O A : O C = O D : O B$ Rearrange the given product into a ratio.
angle $A O D$ = angle $C O B$ Vertically opposite angles are always equal.
triangle $A O D$ is similar to triangle $C O B$ Two sides in the same ratio, with the included angle equal: the SAS criterion.
angle $A$ = angle $C$, angle $D$ = angle $B$ Corresponding angles of similar triangles are equal.
correspondence $A$ to $C$, $O$ to $O$, $D$ to $B$ gives angle $A$ = angle $C$ and angle $D$ = angle $B$ Check both conclusions against the same correspondence. $A$ was matched to $C$ and $D$ to $B$ in the similarity above, so both angle equalities follow from that one match, not two separate arguments.
Two triangles meet where two lines cross, sharing a vertically opposite angle and matching side ratios of 2 to 3.
Check yourself
In the crossing-lines problem, $O A \cdot O B = O C \cdot O D$ rearranges to $O A : O C = O D : O B$, and angle $A O D$ equals angle $C O B$ since they are vertically opposite. Why does this angle equality matter for SAS similarity here?
it proves the two lines $A C$ and $B D$ are parallel
it shows triangle $A O D$ and triangle $C O B$ have equal area
it lets $O A$ be assumed equal to $O D$
it supplies the included angle for the ratio-matched sides
Check your answer
it proves the two lines $A C$ and $B D$ are parallel — Vertically opposite angles being equal says nothing about $A C$ and $B D$ being parallel — they cross at $O$ precisely because they are not.
it shows triangle $A O D$ and triangle $C O B$ have equal area — The equal-angle fact is used to satisfy the angle part of SAS similarity — it says nothing directly about the two triangles’ areas.
it lets $O A$ be assumed equal to $O D$ — Vertically opposite angles being equal is a fact about angles; it gives no information making any two of the four segments equal in length.
✓ it supplies the included angle for the ratio-matched sides — (D) The vertically opposite angle is exactly the one the two ratio-matched sides share.
Lines $E G$ and $F H$ cross at point $O$, with $O E \cdot O H = O F \cdot O G$. Using the same method as $O A \cdot O B = O C \cdot O D$, which triangle pair does SAS similarity establish?
triangle $E O F$ similar to triangle $H O G$, using the vertically opposite angles at $O$
triangle $E O H$ similar to triangle $F O G$, using the same vertically opposite angles at $O$
no similarity can be established without also knowing the four lengths individually
triangle $E O F$ is congruent, not merely similar, to triangle $H O G$
Check your answer
✓ triangle $E O F$ similar to triangle $H O G$, using the vertically opposite angles at $O$ — (A) Rearranging the given product links $E$ with $F$ and $H$ with $G$ through $O$.
triangle $E O H$ similar to triangle $F O G$, using the same vertically opposite angles at $O$ — Rearranging $O E \cdot O H = O F \cdot O G$ to $O E:O F = O G:O H$ links $E$ with $F$ and $G$ with $H$ through $O$, which points to triangle $E O F$ and triangle $H O G$, not $E O H$ and $F O G$.
no similarity can be established without also knowing the four lengths individually — The product relationship, once rearranged into a ratio, is exactly the side-ratio condition SAS similarity needs — no individual lengths are required.
triangle $E O F$ is congruent, not merely similar, to triangle $H O G$ — The given product relationship fixes a ratio between the sides, not their being equal — that supports similarity, not the stronger claim of congruence.
In the crossing-lines figure, a student tries to apply SAS similarity to triangle $A O D$ and triangle $C O B$ using the ratio $O A:O C = O D:O B$, but picks angle $A$ (at vertex $A$, not $O$) as the matched angle instead of the vertically opposite angle at $O$. What is wrong?
angle $A$ is not included between the two ratio-matched sides $O A$ and $O D$
nothing is wrong — any one of the triangle’s three angles can serve equally well as the SAS angle
the ratio should have been written using $O C$ and $O A$ reversed
SAS similarity does not apply to triangles formed by crossing lines
Check your answer
✓ angle $A$ is not included between the two ratio-matched sides $O A$ and $O D$ — (A) The included angle must sit at the vertex shared by both matched sides, which is $O$.
nothing is wrong — any one of the triangle’s three angles can serve equally well as the SAS angle — SAS similarity needs the angle at the vertex shared by the two ratio-matched sides — here that vertex is $O$, not $A$.
the ratio should have been written using $O C$ and $O A$ reversed — The ratio $O A:O C = O D:O B$ is set up correctly; the error is entirely in choosing angle $A$ instead of the included angle at $O$.
SAS similarity does not apply to triangles formed by crossing lines — SAS similarity applies perfectly well here, with the correct included angle at $O$ — the crossing-lines setup is exactly the kind of figure this method solves.
triangle $A B C$ has angles $60^\circ$, $60^\circ$, $60^\circ$ and sides $A B = B C = C A = 4$ cm This is the first triangle, given.
triangle $D E F$ has the same three angles, with sides $D E = E F = F D = 8$ cm Every side is exactly double the first triangle’s.
triangle $A B C$ is similar to triangle $D E F$ The angles match and the sides are in the ratio $1 : 2$, so both similarity conditions hold.
triangle $A B C$ is not congruent to triangle $D E F$ The actual side lengths do not match, $4$ cm against $8$ cm, and no rotation or flip can change that.
triangle $P Q R$ has sides $P Q = Q R = R P = 4$ cm This is a third triangle, built to match triangle $A B C$’s sides exactly.
triangle $A B C$ is congruent to triangle $P Q R$ The two share the same shape and the same size, at scale $1 : 1$.
$A B : D E = 4 : 8$, $B C : E F = 4 : 8$, $C A : F D = 4 : 8$, all $= 1 : 2$ Check the similar pair by confirming all three side ratios agree. They do, and the ratio for the congruent pair, triangle $A B C$ to triangle $P Q R$, comes out $4 : 4 = 1 : 1$ the same way.
Check yourself
Triangle $A B C$ has angles $60^\circ$ each and sides $4$ cm each; triangle $D E F$ has the same three angles but sides $8$ cm each; triangle $P Q R$ has angles $60^\circ$ each and sides $4$ cm each, matching $A B C$ exactly. How should the three triangles be classified in pairs?
all three of the triangles are fully congruent to one another, with no exceptions anywhere at all
$A B C$ and $D E F$ are similar only; $A B C$ and $P Q R$ are congruent (and also similar)
$A B C$ and $D E F$ are congruent; $A B C$ and $P Q R$ are only similar
none of the three triangles are similar to each other, since only two match exactly
Check your answer
all three of the triangles are fully congruent to one another, with no exceptions anywhere at all — $D E F$’s sides are twice as long as $A B C$’s — matching angles alone never overrides that size difference for congruence.
✓ $A B C$ and $D E F$ are similar only; $A B C$ and $P Q R$ are congruent (and also similar) — (B) The doubled sides break congruence for one pair while the matching sides keep it for the other.
$A B C$ and $D E F$ are congruent; $A B C$ and $P Q R$ are only similar — $P Q R$ matches $A B C$’s sides exactly ($4$ cm each), making that pair congruent; $D E F$’s doubled sides make its pair with $A B C$ similar only.
none of the three triangles are similar to each other, since only two match exactly — $A B C$ and $D E F$ share all three angles and a fixed $1:2$ side ratio — that is exactly similarity, even though they are not congruent.
Triangle $J K L$ has angles $45^\circ, 45^\circ, 90^\circ$ and sides $5, 5, 5 \sqrt{2}$ cm. Triangle $M N O$ has the same three angles and sides $10, 10, 10 \sqrt{2}$ cm. Triangle $R S T$ has angles $45^\circ, 45^\circ, 90^\circ$ and sides $5, 5, 5 \sqrt{2}$ cm, matching $J K L$ exactly. Classify the relationship between each pair.
$J K L$ and $M N O$ are similar only; $J K L$ and $R S T$ are congruent
$J K L$ and $M N O$ are congruent, since both are right-angled isosceles triangles
$J K L$ and $R S T$ are only similar, not congruent, since they are two different triangles
none of the three are related, since irrational side lengths cannot be compared
Check your answer
✓ $J K L$ and $M N O$ are similar only; $J K L$ and $R S T$ are congruent — (A) Doubled sides give similarity only; exactly matching sides give congruence.
$J K L$ and $M N O$ are congruent, since both are right-angled isosceles triangles — Being the same type of triangle, right-angled isosceles, does not fix size — $M N O$’s sides are double $J K L$’s, making that pair similar, not congruent.
$J K L$ and $R S T$ are only similar, not congruent, since they are two different triangles — Being drawn as two separate triangles does not prevent congruence — $R S T$’s sides match $J K L$’s exactly, so the pair is congruent as well as similar.
none of the three are related, since irrational side lengths cannot be compared — $5 \sqrt{2}$ is a perfectly definite length, just as comparable as $5$ — the ratios and equalities among the three triangles work the same way whether the numbers are whole or irrational.
A student studies triangle $D E F$ (angles $60^\circ$ each, sides $8$ cm each) and triangle $P Q R$ (angles $60^\circ$ each, sides $4$ cm each) and says: “same angles, so congruent.” What is the accurate classification?
similar
congruent, since matching angles is exactly what congruence tests
neither similar nor congruent, since the side lengths are different
it cannot be decided without knowing which triangle is drawn first
Check your answer
✓ similar — (A) The ratio, not equality, is what the sides actually satisfy here.
congruent, since matching angles is exactly what congruence tests — Congruence needs matching sides as well as angles; here the sides are in a $2:1$ ratio, not equal, so the triangles are similar but not congruent.
neither similar nor congruent, since the side lengths are different — A fixed ratio between sides, $2:1$ here, is exactly what similarity allows — only congruence needs the lengths to be equal, not merely proportional.
it cannot be decided without knowing which triangle is drawn first — The order the two triangles are drawn or named has no bearing on their similarity or congruence — only the angle values and side lengths do.
Congruence is the scale-one case of similarity, so a similarity test with the ratio fixed at 1 turns into a congruence test.
Let us gather everything into five lines.
- SIMILAR: equal angles, proportional sides, any ratio. Triangle $A B C$ and triangle $D E F$ with sides $4$ cm and $8$ cm are similar at ratio $1 : 2$.
- CONGRUENT: similar at ratio $1 : 1$. Two triangles with sides $4$ cm and $4$ cm, and the same angles, are congruent.
- BASIC PROPORTIONALITY: a line parallel to one side divides the other two in the same ratio. $A D : D B = 1.5 : 3$ gives $E C = 2$ cm.
- AA: two matching angles prove similarity. Angle $A$ = angle $D$ = $80^\circ$ is one of the two checked, with no side measured.
- SSS AND SAS: matching side ratios, or one angle plus two side ratios, prove similarity. A $1 : 2$ side ratio found triangle $D E F$’s missing angle at $40^\circ$.
Every criterion here answers the same question. Do two triangles have the same shape, and the same size? AA, SSS and SAS decide similarity. SSS, SAS, ASA and RHS decide congruence. Similarity asks only about shape: equal corresponding angles and proportional corresponding sides. Congruence adds size to the requirement, which is the same as asking for the ratio to be locked at exactly $1 : 1$.
The Basic Proportionality Theorem sits underneath the similarity criteria. A line parallel to one side of a triangle divides the other two sides in that same fixed ratio. Its converse runs the test the other way.
Same shape, any size, is the one question every line above answers a different way.
Check yourself
Which statement best summarises the central question every result in this chapter answers?
when do two triangles have the same area, whatever their shape?
when do two triangles have the same shape, at any size?
when do two triangles have the same perimeter?
when can a triangle be drawn inside another triangle?
Check your answer
when do two triangles have the same area, whatever their shape? — Area is never the chapter’s test — similarity is decided by equal angles and proportional sides, with area playing no role in any criterion.
✓ when do two triangles have the same shape, at any size? — (B) Every criterion in the chapter is a shortcut for answering exactly this one question.
when do two triangles have the same perimeter? — Matching perimeters is not a similarity or congruence test anywhere in this chapter — the tests are always about angles and side ratios.
when can a triangle be drawn inside another triangle? — Whether one triangle fits inside another is a different question from whether two triangles have the same shape — this chapter’s criteria never test containment.
The chapter builds AA, SSS and SAS similarity on top of the Basic Proportionality Theorem, letting a check on just two or three matching parts stand in for all six. What does this buy, practically?
it means every triangle is automatically similar to every other triangle
similarity can be confirmed without measuring every one of a triangle’s six parts
it removes the need to check any angles at all, since only the sides are ever compared
it proves BPT no longer needs its own proof once the criteria exist
Check your answer
it means every triangle is automatically similar to every other triangle — The criteria are tests that some triangle pairs pass and others fail — they do not make every pair of triangles similar.
✓ similarity can be confirmed without measuring every one of a triangle’s six parts — (B) The criteria let a partial check stand in for a full six-part comparison.
it removes the need to check any angles at all, since only the sides are ever compared — AA uses angles only, and SAS uses one angle plus two sides — not every criterion drops angle-checking; the real saving is needing fewer than all six measurements, of either kind.
it proves BPT no longer needs its own proof once the criteria exist — BPT is the foundation the AA, SSS and SAS criteria are built on — the dependency runs from BPT to the criteria, not the other way round.
This chapter treats congruent triangles as similar triangles at scale factor $1:1$. A furniture maker cuts several identical wooden triangular braces from a template. Which idea from this chapter best describes the relationship between any two of these braces?
congruence
similarity only, since they were all cut from the same template
neither similar nor congruent, since they are physical objects, not diagrams
it depends on which brace is measured first
Check your answer
✓ congruence — (A) Matching shape and matching size together is exactly the $1:1$ special case.
similarity only, since they were all cut from the same template — Being cut from the same template gives both matching shape and matching size — that is exactly the $1:1$ scale factor that makes the braces congruent, not merely similar.
neither similar nor congruent, since they are physical objects, not diagrams — Similarity and congruence describe shape and size relationships, which apply just as well to physical triangular braces as to diagrams.
it depends on which brace is measured first — The relationship between two identical braces does not depend on which one happens to be measured first — congruence is a symmetric relationship.
Check yourself: the whole chapter
Triangle $P Q R$ has angles $50^\circ$, $60^\circ$, $70^\circ$. Triangle $X Y Z$ has the same three angles. Every side of $X Y Z$ is exactly double the corresponding side of $P Q R$. Are the two triangles similar?
No — two triangles of different sizes can never be similar, whatever their angles. A fixed doubling like this already rules similarity out.
No — similar triangles need exactly equal side lengths, not merely a fixed ratio between them.
Cannot be decided without measuring the actual side lengths, since equal angles alone say nothing about similarity.
Yes — equal corresponding angles and sides in a fixed ratio (1 : 2 here) are exactly what similarity requires.
Check your answer
No — two triangles of different sizes can never be similar, whatever their angles. A fixed doubling like this already rules similarity out. — Similarity is defined at any ratio between the two triangles’ sizes — a 1 : 2 ratio is just as valid as a 1 : 1 one; only congruence needs the sizes to match exactly.
No — similar triangles need exactly equal side lengths, not merely a fixed ratio between them. — Equal side lengths would make the triangles congruent, a stricter, special case. Similarity only needs the sides in the SAME RATIO, which is exactly what is given here.
Cannot be decided without measuring the actual side lengths, since equal angles alone say nothing about similarity. — The ratio is already fixed and stated, 1 : 2 on every side — nothing further needs measuring to confirm similarity holds.
✓ Yes — equal corresponding angles and sides in a fixed ratio (1 : 2 here) are exactly what similarity requires. — (D) Equal corresponding angles, together with corresponding sides in one fixed ratio, however far from 1 : 1, is exactly the definition of similar triangles.
Triangle $A B C$ is similar to triangle $D E F$, matched in that correspondence order — $A$ with $D$, $B$ with $E$, $C$ with $F$. Which of these must be true?
angle $B$ equals angle $F$
angle $A$ equals angle $E$
angle $B$ equals angle $E$
side $A B$ has no fixed relationship to any side of triangle $D E F$, since the correspondence only fixes matching angles
Check your answer
angle $B$ equals angle $F$ — The stated correspondence pairs B with E, not F — F is matched to C, the third-listed vertex on each side.
angle $A$ equals angle $E$ — The stated correspondence pairs A with D, not E — E is matched to B, the second-listed vertex on each side.
✓ angle $B$ equals angle $E$ — (C) The correspondence order — A with D, B with E, C with F — pairs the second-listed vertex of each triangle, so angle B equals angle E.
side $A B$ has no fixed relationship to any side of triangle $D E F$, since the correspondence only fixes matching angles — Correspondence order fixes BOTH matching angles and matching sides — A B corresponds to D E, and the two are in the same ratio as every other matched side pair.
In triangle $A B C$, a line $D E$ is drawn parallel to side $B C$, with $D$ on side $A B$ and $E$ on side $A C$. By the Basic Proportionality Theorem, which relationship must hold?
$A D : A B = A C : E C$
$A D : D B = A E : E C$
$A D : D B = E C : A E$
$A B : A C = D B : E C$
Check your answer
$A D : A B = A C : E C$ — The theorem compares a side’s two OWN parts to each other — A D : D B — never a part of one side against the whole length of another.
✓ $A D : D B = A E : E C$ — (B) A line parallel to one side of a triangle divides the other two sides in the same ratio — here, A D : D B = A E : E C.
$A D : D B = E C : A E$ — The theorem needs both ratios read in the same direction — A D : D B and A E : E C, both from the vertex A outward — inverting only one side breaks the match.
$A B : A C = D B : E C$ — A B and A C are whole, undivided sides — the theorem is about how D and E split them into parts, A D : D B and A E : E C, not about the whole sides themselves.
Triangle $A B C$ has angle $A = 55^\circ$ and angle $B = 65^\circ$. Triangle $P Q R$ has angle $P = 55^\circ$ and angle $Q = 65^\circ$. Are the two triangles necessarily similar?
No — similarity needs all three pairs of angles checked and confirmed equal, not just two.
No — two triangles can never be judged similar from their angles alone, only from their sides.
Cannot be decided, since the two triangles might still be different sizes. Their sizes would need to be checked before similarity can be confirmed here.
Yes — with two pairs of angles equal, the third pair is automatically equal too, by the angle sum property, so the AA criterion is satisfied.
Check your answer
No — similarity needs all three pairs of angles checked and confirmed equal, not just two. — The third angle never needs a separate check — once two pairs of angles match, the third pair is forced to match too, since every triangle’s angles sum to 180 degree.
No — two triangles can never be judged similar from their angles alone, only from their sides. — Angles alone are exactly what the AA criterion uses — two matching angle pairs are already enough to guarantee similarity, without checking a single side.
Cannot be decided, since the two triangles might still be different sizes. Their sizes would need to be checked before similarity can be confirmed here. — Similarity holds at any size ratio between the two triangles — differing sizes are exactly what separates similar triangles from congruent ones, not a reason to doubt similarity.
✓ Yes — with two pairs of angles equal, the third pair is automatically equal too, by the angle sum property, so the AA criterion is satisfied. — (D) Two matching angle pairs force the third to match as well, by the angle sum property — that is exactly what the AA criterion guarantees.
Triangle $A B C$ has sides $A B = 3$, $B C = 4$, $C A = 5$. Triangle $D E F$ has sides $D E = 6$, $E F = 8$, $F D = 10$. Using the SSS similarity criterion, are the two triangles similar?
No — the two triangles have different perimeters, so they cannot be similar.
Yes — all three pairs of corresponding sides are in the same ratio, 1 : 2, so SSS similarity applies.
No — SSS similarity needs the corresponding sides to be equal in length, not merely proportional. Only equal side lengths satisfy this test.
Cannot be decided without also checking whether the corresponding angles are equal.
Check your answer
No — the two triangles have different perimeters, so they cannot be similar. — Perimeters scale along with every side, so a different perimeter is exactly what a 1 : 2 ratio predicts — it says nothing against similarity.
✓ Yes — all three pairs of corresponding sides are in the same ratio, 1 : 2, so SSS similarity applies. — (B) $3/6 = 4/8 = 5/10 = 1/2$ — all three side ratios match, so by SSS similarity the two triangles are similar.
No — SSS similarity needs the corresponding sides to be equal in length, not merely proportional. Only equal side lengths satisfy this test. — Equal side lengths would test for congruence, a stricter, special case. SSS similarity only needs the three sides in the SAME RATIO, which is exactly what holds here, 1 : 2 throughout.
Cannot be decided without also checking whether the corresponding angles are equal. — The SSS criterion’s whole point is that matching all three side ratios is already enough — the angles are then guaranteed to match too, with no separate check needed.
A shortcut D E parallel to the road cuts the two trails in the same ratio, 1.2 to 1.8 and 1.6 to 2.4.
You will meet similar and congruent triangles often. A shape gets scaled to a new size. Two pieces must match exactly. Here are seven of those places.
Sewing a matching pair of pennant flags. A flag maker sews a small triangular pennant with sides $30$, $40$, $50$ cm. A large pennant for the same client must be similar, at scale factor $2$. The maths: equal angles and sides in the same ratio make two triangles similar, whatever the size. Large sides $60$, $80$, $100$ cm: $60/30 = 2$, $80/40 = 2$, $100/50 = 2$, one ratio throughout.
Scaling up a triangular rug pattern. A paper pattern piece has sides $5$, $6$, $9$ cm. The rug maker cuts a piece at scale factor $2$ and wants to check it was cut correctly. The maths: all three pairs of sides are in the same ratio. That alone makes the two triangles similar, with no angle measured. Cut sides $10$, $12$, $18$ cm: $10/5 = 2$, $12/6 = 2$, $18/9 = 2$. The cut is correct.
Checking two garden beds need equal edging. One triangular flower bed has sides $3$, $4$, $5$ m. Another has sides $5$, $4$, $3$ m, measured in a different order. The maths: all three sides of one triangle equal the three sides of another. That makes the triangles congruent, so they need the same edging length. Both beds: $3 + 4 + 5 = 12$ m of edging, since the two lists of sides are the same three numbers.
Finding a distance along a shortcut trail. Two trails leave a gate $A$ and reach the main road at $B$ and $C$. A shortcut parallel to the road crosses them at $D$ and $E$. The maths: a line parallel to one side of a triangle divides the other two sides in the same ratio. From the start, $A D = 1.2$ km and $D B = 1.8$ km. Also $A E = 1.6$ km, so $A D : D B = A E : E C$ gives $E C = 2.4$ km.
Cutting a sandwich exactly in half. A rectangular sandwich, $12$ cm by $8$ cm, is cut once along the diagonal into two triangles. The maths: each half has sides $12$ and $8$ cm with a right angle between them. So SAS makes the two halves congruent. Each triangle has legs $12$ and $8$ cm and area $48$ square cm. Together, $48 + 48 = 96$, the whole sandwich.
A diagonal always splits a rectangle into two congruent triangles, so each half carries exactly half the area.
Checking two sailboat sails are the same shape. A sailing club’s junior sail and senior sail are both triangular. Both have base angles $70^\circ$ and $60^\circ$. The maths: two matching angles are enough to prove similarity. That is AA. The larger sail behaves the same way, just bigger. The third angle is fixed too: $180 - 70 - 60 = 50^\circ$, in both sails.
Buying a set square in two sizes. A stationery shop sells a $30^\circ$-$60^\circ$-$90^\circ$ set square in two sizes, junior and senior. The maths: the two set squares share the same three angles, so they are similar. Their sides are not equal, so they are not congruent. The junior size has hypotenuse $10$ cm. The senior size has hypotenuse $20$ cm. Both keep angles $30^\circ$, $60^\circ$, $90^\circ$.
Your turn. Two triangular kite frames have sides $6$, $8$, $10$ cm and $9$, $12$, $15$ cm. Are they similar? What is the scale factor? Answer: Check the ratios: $9/6 = 1.5$, $12/8 = 1.5$, $15/10 = 1.5$. All equal, so the kites are similar, at scale factor $1.5$.
practice Are any two regular hexagons similar? Give a reason. (Worked in full below — read it, then do the next two the same way.)
practice Two right triangles each have an angle of $30^\circ$. Are they similar? Give a reason. (Start the same way — write down all three angles of each triangle before deciding.)
practice One isosceles triangle has a vertex angle of $40^\circ$ and another has a vertex angle of $80^\circ$. Are they similar? Give a reason. (Work out the two base angles of each first.)
practice For each pair below, state whether they are always similar, never similar, or only sometimes similar, and give a reason. (a) Any two circles. (b) Any two squares. (c) Any two equilateral triangles. (d) Any two rectangles.
practice Give one example of two figures that are similar, and one example of two figures that are not similar. Give a reason for each.
practice A quadrilateral has all four angles equal to $90^\circ$. A second quadrilateral also has all four angles equal to $90^\circ$, but its sides are not in the same ratio as the first. Are the two quadrilaterals similar? Give a reason.
practice Two rhombuses have all four sides in the ratio $1 : 1 : 1 : 1$ to each other, but one has an angle of $70^\circ$ and the other has an angle of $100^\circ$. Are the two rhombuses similar? Give a reason.
practice State whether every pair of squares is similar, and why.
Answers
Yes, always. Every angle is $120^\circ$ in both, and all six sides of a regular hexagon are equal, so corresponding sides are in the same ratio.
Yes. Each has angles $90^\circ$, $30^\circ$ and $60^\circ$, so all three angles match.
No. The first has angles $40^\circ$, $70^\circ$, $70^\circ$ and the second has $80^\circ$, $50^\circ$, $50^\circ$, so the angles do not match.
(a) Always similar — every circle is the same shape, whatever its radius. (b) Always similar — every square has four $90^\circ$ angles and all sides equal. (c) Always similar — every angle is fixed at $60^\circ$. (d) Only sometimes similar — angles always match at $90^\circ$, but sides need not be in the same ratio.
Sample answer: any two circles are similar (equal angles are not needed; every circle has the same shape at some ratio). A square and a non-square rectangle are not similar — equal angles, but sides not in the same ratio.
Not similar — the angles match, but the sides are not in the same ratio; both conditions are needed together.
Not similar — the sides are in the same ratio, but the angles differ.
Yes — every square has four $90^\circ$ angles and all sides equal, so any two squares always satisfy both similarity conditions.
Exercise 6.1 — further practice
practice Are all isosceles triangles always similar to each other? Give a reason.
practice An equilateral triangle has side $3$ cm and a second equilateral triangle has side $9$ cm. Are the two triangles similar? Give a reason.
practice A photograph of size $10$ cm by $15$ cm is enlarged to $20$ cm by $25$ cm. Are the original and the enlarged photograph similar rectangles? Give a reason.
practice Which one of the following pairs is NOT always similar?
Two squares
Two equilateral triangles
Two isosceles triangles
Two circles
practice One quadrilateral has all four angles equal to $90^\circ$. A second quadrilateral has angles $100^\circ$, $80^\circ$, $100^\circ$, $80^\circ$, in order. Are the two quadrilaterals similar? Give a reason.
Answers
No — an isosceles triangle with angles $50^\circ$, $50^\circ$, $80^\circ$ and one with angles $40^\circ$, $40^\circ$, $100^\circ$ are both isosceles, but their angles differ, so they are not similar.
Yes — every equilateral triangle has all three angles equal to $60^\circ$, whatever its side length, so the angle condition always holds.
No — the ratio of the widths is $10 : 20 = 1 : 2$ and the ratio of the heights is $15 : 25 = 3 : 5$. The two ratios are not equal, so the sides are not in the same ratio, and the rectangles are not similar.
C — Two isosceles triangles.
No — the corresponding angles are not equal, so the first condition for similarity already fails; the sides need not even be compared.
practice In triangle $A B C$, $D$ lies on $A B$ and $E$ lies on $A C$, with $D E$ parallel to $B C$. If $A D = 2$ cm, $D B = 5$ cm and $A E = 3$ cm, find $E C$. (Worked in full below — read it, then do the next two the same way.)
practice In triangle $P Q R$, $S$ lies on $P Q$ and $T$ lies on $P R$, with $P S = 2$ cm, $S Q = 3$ cm, $P T = 3$ cm and $T R = 4.5$ cm. Is $S T$ parallel to $Q R$? (Start the same way, but here both ratios are known — work them out and compare.)
practice In triangle $A B C$, a line parallel to $B C$ meets $A B$ at $D$ and $A C$ at $E$, with $A D : D B = 2 : 3$. If $A E = 4$ cm, find $E C$. (The ratio is handed to you this time — it goes straight into the left-hand side.)
practice In triangle $A B C$, $D$ lies on $A B$ and $E$ lies on $A C$, with $D E$ parallel to $B C$. If $A D = 4$ cm, $D B = 6$ cm and $A E = 6$ cm, find $E C$.
practice In triangle $P Q R$, $S$ lies on $P Q$ and $T$ lies on $P R$, with $P S = 4$ cm, $S Q = 4.5$ cm, $P T = 8$ cm, $T R = 9$ cm. Is $S T$ parallel to $Q R$?
practice In triangle $A B C$, $D$ and $E$ are the midpoints of $A B$ and $A C$. Prove that $D E$ is parallel to $B C$.
practice $A B C D$ is a trapezium with $A B$ parallel to $D C$. The diagonals $A C$ and $B D$ meet at $O$. Prove that $A O : O C = B O : O D$.
practice In triangle $A B C$, a line parallel to $B C$ meets $A B$ at $D$ and $A C$ at $E$, with $A D = 2 D B$. If $A E = 6$ cm, find $E C$.
Answers
$E C = 7.5$ cm.
Yes. $P S : S Q = 2 : 3$ and $P T : T R = 3 : 4.5 = 2 : 3$. The two ratios are equal, so by the converse of the theorem $S T$ is parallel to $Q R$.
$E C = 6$ cm.
$A D : D B = A E : E C$, so $4 : 6 = 6 : E C$, giving $E C = 9$ cm.
$P S : S Q = 4 : 4.5 = 8 : 9$ and $P T : T R = 8 : 9$. The two ratios are equal, so by the converse of the Basic Proportionality Theorem, $S T$ is parallel to $Q R$.
$D$ and $E$ are midpoints, so $A D : D B = 1 : 1 = A E : E C$. Equal ratios on both sides means, by the converse of the Basic Proportionality Theorem, $D E$ is parallel to $B C$.
Since $A B$ is parallel to $D C$, angle $O A B$ = angle $O C D$ and angle $O B A$ = angle $O D C$ (alternate angles). So triangle $A O B$ is similar to triangle $C O D$ by AA, giving $A O : O C = B O : O D$.
$A D = 2 D B$ means $A D : D B = 2 : 1$. By the Basic Proportionality Theorem, $A E : E C = 2 : 1$ too, so $E C = 3$ cm.
Exercise 6.2 — further practice
practice In triangle $A B C$, $D$ lies on $A B$ and $E$ lies on $A C$, with $D E$ parallel to $B C$. If $A D = 2$ cm, $D B = 3$ cm and $A E = 2.4$ cm, find $E C$.
practice In triangle $P Q R$, $S$ lies on $P Q$ and $T$ lies on $P R$, with $S T$ parallel to $Q R$. If $P S = 3$ cm, $S Q = 5$ cm and $P T = 6$ cm, find $T R$.
practice In triangle $P Q R$, $E$ lies on $P Q$ and $F$ lies on $P R$, with $P E = 3$ cm, $E Q = 4.5$ cm, $P F = 4$ cm and $F R = 5$ cm. Which statement is correct?
$E F$ is parallel to $Q R$
$E F$ is not parallel to $Q R$
$E F$ and $Q R$ cannot be compared without more information
$E F$ is perpendicular to $Q R$
practice A surveyor marks points $D$ on side $A B$ and $E$ on side $A C$ of a triangular plot $A B C$, to check whether a proposed fence $D E$ can run parallel to $B C$. She measures $A D = 6$ m, $D B = 9$ m, $A E = 8$ m and $E C = 12$ m. Can the fence $D E$ be drawn parallel to $B C$? Give a reason.
practice In triangle $A B C$, $D$ lies on $A B$ and $E$ lies on $A C$, with $D E$ parallel to $B C$. If $A D : D B = 3 : 4$ and $A E = 6$ cm, find $E C$.
practice A carpenter fixes a beam $D E$ parallel to the base $B C$ inside a triangular truss $A B C$, with $D$ on $A B$ and $E$ on $A C$. If $A D = 2.5$ m, $D B = 5$ m and $A E = 3$ m, find $E C$.
practice $A B C D$ is a trapezium with $A B$ parallel to $D C$. The diagonals $A C$ and $B D$ meet at $O$, so that $A O : O C = B O : O D$. If $A O = 3$ cm, $B O = 4.5$ cm and $O D = 6$ cm, find $O C$.
practice In triangle $A B C$, $D$, $E$ and $F$ are the midpoints of sides $B C$, $C A$ and $A B$. Show that triangle $D E F$ is similar to triangle $A B C$, and state the scale factor.
practice $A B C D$ is a trapezium with $A B$ parallel to $D C$. $E$ lies on $A D$ and $F$ lies on $B C$, with $E F$ parallel to $A B$. Prove that $A E : E D = B F : F C$.
Answers
$E C = 3.6$ cm
$T R = 10$ cm
B — $E F$ is not parallel to $Q R$.
Yes — $A D : D B = 6 : 9 = 2 : 3$ and $A E : E C = 8 : 12 = 2 : 3$. The two ratios are equal, so by the converse of the Basic Proportionality Theorem, $D E$ can be drawn parallel to $B C$.
$E C = 8$ cm
$E C = 6$ m
$O C = 4$ cm
Since $E$ and $F$ are midpoints of $C A$ and $A B$, segment $E F$ is parallel to $B C$ and $E F = 1/2 B C$ by the midpoint theorem. Likewise $D E = 1/2 A B$ and $F D = 1/2 C A$. All three sides of triangle $D E F$ are exactly half the corresponding sides of triangle $A B C$ ($D$ opposite $A$, $E$ opposite $B$, $F$ opposite $C$), so triangle $D E F$ is similar to triangle $A B C$ by SSS, with scale factor $1 : 2$.
Join the diagonal $A C$, meeting $E F$ at $G$. In triangle $A D C$, $E G$ is parallel to $D C$ (both parallel to $A B$), so by the Basic Proportionality Theorem $A E : E D = A G : G C$. In triangle $A B C$, $G F$ is parallel to $A B$, so by the Basic Proportionality Theorem $C G : G A = C F : F B$, which gives $A G : G C = B F : F C$. Combining the two, $A E : E D = B F : F C$.
practice Triangle $A B C$ has angle $A = 40^\circ$ and angle $B = 75^\circ$. Triangle $P Q R$ has angle $P = 40^\circ$ and angle $Q = 75^\circ$. State whether the two triangles are similar and name the criterion. (Worked in full below — read it, then do the next two the same way.)
practice Triangle $L M N$ has sides $L M = 4$ cm, $M N = 6$ cm and $N L = 8$ cm. Triangle $X Y Z$ has sides $X Y = 6$ cm, $Y Z = 9$ cm and $Z X = 12$ cm. Show that the two triangles are similar and name the criterion. (Start the same way, but here it is three side ratios that must be compared.)
practice In triangle $A B C$, $A B = 4$ cm, $A C = 6$ cm and angle $A = 60^\circ$. In triangle $P Q R$, $P Q = 6$ cm, $P R = 9$ cm and angle $P = 60^\circ$. Are the two triangles similar? Name the criterion. (Two sides and the angle between them — check the two ratios, then the angle.)
practice Triangle $A B C$ has angle $A = 50^\circ$ and angle $B = 70^\circ$. Triangle $D E F$ has angle $D = 50^\circ$ and angle $E = 70^\circ$. State, with reason, whether the two triangles are similar.
practice Triangle $P Q R$ has sides $P Q = 3$ cm, $Q R = 4$ cm, $R P = 5$ cm. Triangle $X Y Z$ has sides $X Y = 6$ cm, $Y Z = 8$ cm, $Z X = 10$ cm. Show that the two triangles are similar, and name the criterion used.
practice In triangle $A B C$, angle $A = 40^\circ$; in triangle $D E F$, angle $D = 40^\circ$. If $A B : D E = A C : D F = 2 : 3$, are the two triangles similar? Name the criterion.
practice Triangle $A B C$ is similar to triangle $D E F$, with angle $A = 65^\circ$ and angle $B = 55^\circ$. Find angle $F$.
practice $A D$ is the altitude from $A$ to $B C$ in a right triangle $A B C$, right-angled at $A$. Show that triangle $A B D$ is similar to triangle $C B A$.
Answers
Yes, by AA. Two pairs of angles are equal, and the third pair follows, since the angles of a triangle add to $180^\circ$.
$L M : X Y = 4 : 6$, $M N : Y Z = 6 : 9$ and $N L : Z X = 8 : 12$ are all $2 : 3$, so the two triangles are similar by SSS.
Yes, by SAS. $A B : P Q = 4 : 6$ and $A C : P R = 6 : 9$ are both $2 : 3$, and the included angles $A$ and $P$ are both $60^\circ$.
Yes, similar by AA — angle $A$ = angle $D$ = $50^\circ$ and angle $B$ = angle $E$ = $70^\circ$, matching in correspondence order.
$P Q : X Y = 3 : 6$, $Q R : Y Z = 4 : 8$, $R P : Z X = 5 : 10$ all equal $1 : 2$, so the triangles are similar by SSS.
Yes, similar by SAS — one matching angle, with the two including sides in the same ratio.
Angle $C = 180^\circ - 65^\circ - 55^\circ = 60^\circ$; equal to angle $F$, so angle $F = 60^\circ$.
Triangle $A B D$ and triangle $C B A$ share angle $B$, and angle $A D B$ = angle $B A C$ = $90^\circ$ (both right angles). Two matching angles are enough — the two triangles are similar by AA.
Exercise 6.3 — further practice
practice Triangle $A B C$ has angle $A = 55^\circ$ and angle $B = 50^\circ$. Triangle $D E F$ has angle $D = 55^\circ$ and angle $F = 75^\circ$. Are the two triangles similar? Name the criterion, and state the correspondence.
practice Triangle $A B C$ has sides $4$ cm, $5$ cm and $6$ cm. Triangle $P Q R$ has sides $8$ cm, $10$ cm and $12$ cm. Triangle $X Y Z$ has sides $8$ cm, $9$ cm and $10$ cm. Which one of triangle $P Q R$ and triangle $X Y Z$ is similar to triangle $A B C$? Name the criterion.
practice Triangle $A B C$ has $A B = 5$ cm, $A C = 7$ cm and angle $A = 40^\circ$. Triangle $D E F$ has $D E = 10$ cm, $D F = 14$ cm and angle $D = 40^\circ$. Are the two triangles similar? Name the criterion.
practice Triangle $A B C$ is similar to triangle $P Q R$, with angle $A = 80^\circ$ and angle $C = 45^\circ$. Find angle $Q$.
practice Triangle $A B C$ is similar to triangle $D E F$, with $A B = 6$ cm, $D E = 9$ cm and $B C = 8$ cm. Find $E F$.
practice Triangle $A B C$ is similar to triangle $D E F$, with correspondence $A$ to $D$, $B$ to $E$, $C$ to $F$. Which of these ratios must equal $A B : D E$?
$B C : E F$
$B C : D E$
$A C : D E$
$D E : A B$
practice Triangle $A B C$ is right-angled at $B$. $M$ is a point on ray $A B$ and $P$ is a point on ray $A C$, so that triangle $A M P$ is right-angled at $M$. Show that triangle $A B C$ is similar to triangle $A M P$, and that $C A : P A = B C : M P$.
practice In triangle $A B C$, $D$ is the midpoint of $B C$, and $E$ lies on $A C$ so that $D E$ is parallel to $A B$. Show that triangle $C D E$ is similar to triangle $C B A$, and find the ratio $C D : C B$.
practice In triangle $A B C$, $D$ lies on $A B$ and $E$ lies on $A C$, with angle $A D E$ = angle $A C B$. Show that triangle $A D E$ is similar to triangle $A C B$.
practice In triangle $A B C$, $A D$ is the median to $B C$. In triangle $P Q R$, $P M$ is the median to $Q R$. Given that $A B : P Q = B C : Q R = A D : P M$, show that triangle $A B C$ is similar to triangle $P Q R$.
practice A vertical pole $4$ m tall casts a shadow $3$ m long on the ground. At the same time, a nearby tower casts a shadow $15$ m long. Find the height of the tower.
Answers
Yes — angle $C = 180^\circ - 55^\circ - 50^\circ = 75^\circ$, and angle $E = 180^\circ - 55^\circ - 75^\circ = 50^\circ$. So angle $A$ = angle $D$, angle $B$ = angle $E$ and angle $C$ = angle $F$, and the two triangles are similar by AA, with correspondence $A$ to $D$, $B$ to $E$, $C$ to $F$.
Triangle $P Q R$ — its sides are each exactly twice the matching side of triangle $A B C$ ($4 : 8 = 5 : 10 = 6 : 12 = 1 : 2$), so the two are similar by SSS. Triangle $X Y Z$ is not similar to triangle $A B C$: the ratios $4 : 8$, $5 : 9$ and $6 : 10$ are not all equal.
Yes — $A B : D E = 5 : 10 = 1 : 2$ and $A C : D F = 7 : 14 = 1 : 2$, with the included angle $A$ = angle $D$ = $40^\circ$. The two triangles are similar by SAS.
angle $Q = 55^\circ$
$E F = 12$ cm
A — $B C : E F$.
Angle $A$ is common to both triangles, since $M$ lies on ray $A B$ and $P$ lies on ray $A C$. Angle $B$ = angle $M$ = $90^\circ$. Two matching angles are enough, so triangle $A B C$ is similar to triangle $A M P$ by AA, with correspondence $A$ to $A$, $B$ to $M$, $C$ to $P$. Corresponding sides are then in the same ratio, so $C A : P A = B C : M P$.
Angle $C$ is common to both triangles. Since $D E$ is parallel to $A B$, angle $C D E$ = angle $C B A$ (corresponding angles). Two matching angles are enough, so triangle $C D E$ is similar to triangle $C B A$ by AA. Since $D$ is the midpoint of $B C$, $C D : C B = 1 : 2$.
Angle $A$ is common to both triangles, since angle $D A E$ and angle $C A B$ are the same angle. Angle $A D E$ = angle $A C B$ is given. Two matching angles are enough, so triangle $A D E$ is similar to triangle $A C B$ by AA, with correspondence $A$ to $A$, $D$ to $C$, $E$ to $B$.
Since $A D$ and $P M$ are medians, $B D = 1/2 B C$ and $Q M = 1/2 Q R$, so $B D : Q M = B C : Q R$ too. Triangle $A B D$ has all three sides ($A B$, $B D$, $A D$) in the same ratio as the matching sides ($P Q$, $Q M$, $P M$) of triangle $P Q M$, so triangle $A B D$ is similar to triangle $P Q M$ by SSS. This gives angle $B$ = angle $Q$. In triangles $A B C$ and $P Q R$, $A B : P Q = B C : Q R$ is given, and the included angle $B$ = angle $Q$ has just been shown, so triangle $A B C$ is similar to triangle $P Q R$ by SAS.
