Lesson 3 ~40 min Geometrical Figures C · Path +85 XP

Formal Similarity Proofs

Similarity is congruence with the size requirement removed. That single change turns three of the four congruence tests into similarity tests, and turns every conclusion from an equality into a ratio.

Today's hook: You cannot measure the height of a tree with a ruler. But stand a metre stick beside it, measure both shadows, and two similar triangles hand you the answer. Similarity is how mathematics reaches things it cannot touch, and a formal proof is what makes the answer trustworthy.

0/5QUESTS

Think First

warm-up

Two triangles have angles $40°$, $60°$, $80°$ and $40°$, $60°$, $80°$. Must they be the same size? Must they be the same shape? Now suppose you are told one side of the first is $3$ cm and the matching side of the second is $9$ cm. What can you now say about every other pair of matching sides, and why?

Record your answer in your workbook.

The Big Idea

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Two triangles are similar when their angles match and their matching sides are in a constant ratio. The symbol is $\sim$, and the order of the letters carries the correspondence exactly as it does for congruence. Because size is no longer fixed, the conclusions you draw are ratios, not equalities.

$$\triangle ABC \sim \triangle PQR \quad\Longrightarrow\quad \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$$

That constant is the scale factor. One ratio establishes it, and every other pair of matching sides must then obey it, which is what makes similarity useful: measure what you can reach, and the ratio delivers what you cannot. Note that congruence is simply the case where the scale factor is $1$.

$\triangle ABC \sim \triangle PQR$

Two angles are enough

The third follows from the angle sum, so AA is a complete test.

Ratios, not equalities

Similar triangles give $\frac{AB}{PQ} = \frac{BC}{QR}$, never $AB = PQ$.

Order still matters

Write $\sim$ with matching vertices, or the ratios pair the wrong sides.

What You'll Master

objectives

Know

The three similarity tests: equiangular (AA), three sides in ratio (SSS), and two sides in ratio with the included angle (SAS)
That similar triangles have matching sides in a constant ratio, the scale factor
That congruence is the special case of similarity with scale factor $1$

Understand

Why two equal angles are sufficient, where congruence needed a side as well
Why the conclusion of a similarity proof is a ratio rather than an equality

Can Do

Set out a formal similarity proof with a reason on every line
Use a proven similarity to calculate an unknown length
Recognise similar triangles inside a single figure, including the overlapping case

Words You Need

vocabulary

SimilarSame shape: matching angles equal and matching sides in a constant ratio. Written with the symbol $\sim$.

Scale factorThe constant ratio between matching sides of two similar figures.

EquiangularHaving all matching angles equal. For triangles this alone proves similarity.

AA testTwo pairs of equal angles, which forces the third and so proves similarity.

Matching sidesSides in corresponding positions, identified by the order of letters in the similarity statement.

TransversalA line crossing two others, creating the equal angles that similarity proofs often use.

The Three Similarity Tests

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Take the congruence tests and remove the requirement that lengths be equal, replacing it with a requirement that they be in ratio.

Test	What you need	Congruence version
AA (equiangular)	two pairs of equal angles	none; AAA proves nothing about congruence
SSS	all three pairs of sides in the same ratio	all three pairs equal
SAS	two pairs of sides in ratio, with equal included angles	two pairs equal, with equal included angle

The striking row is the first. For congruence, three equal angles are worthless; for similarity they are everything. That is the whole difference between the two ideas, stated in one line: angles determine shape, and similarity is a claim about shape alone.

Two angles suffice, because the third is forced by the angle sum. So the test is normally quoted as AA rather than AAA.

Setting Out a Similarity Proof

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The form is identical to a congruence proof. Establish the facts a test needs, name the test, write the similarity statement in matching order, then draw conclusions.

Statement	Reason
$\angle A = \angle P$	given
$\angle B = \angle Q$	given
$\triangle ABC \sim \triangle PQR$	equiangular (AA)
$\dfrac{AB}{PQ} = \dfrac{BC}{QR}$	matching sides of similar triangles are in ratio

Notice the final reason. In a congruence proof it was "matching sides of congruent triangles are equal"; here it is "in ratio". Writing an equality instead of a ratio is the single most common error in similarity work, and it usually comes from copying the habit across from congruence.

Using Similarity to Find a Length

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Once the similarity is proved, one known pair of matching sides fixes the scale factor, and every other pair must obey it.

Suppose $\triangle ABC \sim \triangle PQR$, with $AB = 6$, $PQ = 9$ and $BC = 8$. The scale factor from the first triangle to the second is

$$k = \frac{PQ}{AB} = \frac{9}{6} = \frac{3}{2}$$

so $QR = k \times BC = \dfrac{3}{2} \times 8 = 12$.

Equivalently, set up the ratio and solve:

$$\frac{AB}{PQ} = \frac{BC}{QR} \quad\Longrightarrow\quad \frac{6}{9} = \frac{8}{QR} \quad\Longrightarrow\quad QR = \frac{9 \times 8}{6} = 12$$

Sanity check

Decide first whether the answer should be bigger or smaller. Here the second triangle is larger, so $QR$ must exceed $BC = 8$. An answer of $5.3$ would signal an inverted ratio.

Similar Triangles Hidden in One Figure

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Most exam questions do not present two separate triangles. They present one figure containing two overlapping triangles, and finding them is half the work.

The parallel-line figure. If $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$, then $\triangle ADE$ and $\triangle ABC$ share the angle at $A$, and the parallel lines give $\angle ADE = \angle ABC$ as corresponding angles. Two angles, so $\triangle ADE \sim \triangle ABC$ by AA.

The altitude figure. In a right-angled triangle, the altitude to the hypotenuse creates two smaller triangles, each similar to the original and to each other, because each shares an angle with the original and has its own right angle.

In both cases the technique is the same: look for a shared angle, then find one more from parallel lines or a right angle. That is AA, and it is enough.

Common Pitfalls

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Concluding $AB = PQ$ from a similarity instead of $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$.

Fix: similar triangles are the same shape, not the same size. Only when the scale factor is $1$ do the sides become equal, and that is congruence.

Pairing sides from the diagram's layout rather than from the similarity statement, so the ratio compares the wrong two sides.

Fix: read the letters. In $\triangle ABC \sim \triangle PQR$, $AB$ matches $PQ$ because $A$ matches $P$ and $B$ matches $Q$. The picture's arrangement is irrelevant.

Inverting the ratio, so a length comes out smaller when it should be larger.

Fix: predict the direction before calculating. If the second triangle is the larger, every one of its sides must exceed its partner.

Watch Me Solve It · 3 examples

Watch Me Solve It · A proof using parallel lines

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PROBLEM

In $\triangle ABC$, the point $D$ lies on $AB$ and $E$ lies on $AC$, with $DE \parallel BC$. Prove that $\triangle ADE \sim \triangle ABC$.

1

Find the shared angle

$\angle DAE = \angle BAC \quad \text{(common angle at } A)$

The two triangles overlap at $A$, so that angle belongs to both.
2

Use the parallel lines

$\angle ADE = \angle ABC \quad \text{(corresponding angles, } DE \parallel BC)$

The parallel marks are given information.
3

Name the test

$\triangle ADE \sim \triangle ABC \quad \text{(equiangular, AA)}$

Two pairs of equal angles force the third, so no more is needed.
4

State what now follows

$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$

Matching sides of similar triangles are in ratio, read off the order of the letters.

Answer$\triangle ADE \sim \triangle ABC$ by AA

Watch Me Solve It · Finding an unknown length

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PROBLEM

$\triangle PQR \sim \triangle STU$, with $PQ = 12$ cm, $ST = 8$ cm and $QR = 15$ cm. Find $TU$.

1

Identify the matching pairs from the statement

$PQ \leftrightarrow ST, \qquad QR \leftrightarrow TU$

$P$ matches $S$, $Q$ matches $T$, $R$ matches $U$.
2

Predict the direction

$ST < PQ \Rightarrow \text{the second triangle is smaller}$

So $TU$ must be less than $QR = 15$. Predicting first catches an inverted ratio.
3

Write the ratio and substitute

$\frac{PQ}{ST} = \frac{QR}{TU}$

$\frac{12}{8} = \frac{15}{TU}$
4

Solve and check against the prediction

$TU = \frac{8 \times 15}{12} = 10$

$10 < 15$, as predicted.

Answer$TU = 10$ cm

Watch Me Solve It · Proving similarity from sides in ratio

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PROBLEM

$\triangle ABC$ has sides $AB = 4$, $BC = 6$, $CA = 8$. $\triangle PQR$ has $PQ = 6$, $QR = 9$, $RP = 12$. Prove the triangles are similar and state the scale factor.

1

Pair the sides in matching order

$\frac{PQ}{AB} = \frac{6}{4}, \quad \frac{QR}{BC} = \frac{9}{6}, \quad \frac{RP}{CA} = \frac{12}{8}$

Each pair is read from the intended correspondence $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$.
2

Simplify each ratio

$\frac{6}{4} = \frac{3}{2}, \quad \frac{9}{6} = \frac{3}{2}, \quad \frac{12}{8} = \frac{3}{2}$

All three must reduce to the same value, or the triangles are not similar.
3

Name the test

$\triangle ABC \sim \triangle PQR \quad \text{(three pairs of sides in the same ratio, SSS)}$
4

State the scale factor

$k = \frac{3}{2}$

From $\triangle ABC$ to $\triangle PQR$. Going the other way the factor is $\dfrac{2}{3}$, so say which direction you mean.

AnswerSimilar by SSS, with scale factor $\dfrac{3}{2}$

Brain Trainer · 4 problems

Brain Trainer · Test, ratio and length

4 problems

Four quick problems. Work each one, then reveal the answer.

1 Two triangles have two pairs of equal angles. Which similarity test applies?

The third angle follows from the angle sum, so two are enough.AA (equiangular)
2 $\triangle ABC \sim \triangle PQR$. Which side matches $CA$?

$C$ matches $R$ and $A$ matches $P$.$RP$
3 Similar triangles have matching sides $5$ cm and $20$ cm. What is the scale factor from the smaller to the larger?

$20 \div 5$.$4$
4 $\triangle ABC \sim \triangle PQR$ with $AB = 3$, $PQ = 12$, $BC = 5$. Find $QR$.

The scale factor is $4$, and $QR$ matches $BC$.$20$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

The AA test

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To prove two triangles similar by the equiangular test, you need:

MC2

What similarity gives you

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If $\triangle ABC \sim \triangle PQR$, then:

MC3

Congruence and similarity

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Two congruent triangles are:

MC4

Reading the ratio

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$\triangle ABC \sim \triangle PQR$, with $AB = 5$, $PQ = 15$ and $CA = 7$. The length $RP$ is:

MC5

Overlapping triangles

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In $\triangle ABC$, $D$ is on $AB$ and $E$ is on $AC$ with $DE \parallel BC$. The two angles that prove $\triangle ADE \sim \triangle ABC$ are:

Short Answer · 3 questions

Prove and calculate

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SHORT ANSWER

In $\triangle PQR$, the point $S$ lies on $PQ$ and $T$ lies on $PR$, with $ST \parallel QR$. Also $PS = 4$ cm, $SQ = 6$ cm and $ST = 5$ cm.
(a) Prove that $\triangle PST \sim \triangle PQR$.
(b) Find the scale factor from $\triangle PST$ to $\triangle PQR$.
(c) Find the length of $QR$.

Write your working in your book.

Which test, and what follows

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SHORT ANSWER

For each pair, state whether similarity can be proved, name the test if so, and give one conclusion that follows.
(a) $\triangle ABC$ and $\triangle PQR$ with $\angle A = \angle P = 50°$ and $\angle B = \angle Q = 60°$
(b) $\triangle ABC$ with sides $3, 5, 7$ and $\triangle PQR$ with sides $9, 15, 21$
(c) $\triangle ABC$ and $\triangle PQR$ with $AB = 4$, $PQ = 8$, $BC = 5$, $QR = 10$ and $\angle B = \angle Q$

Write your working in your book.

Find the fault

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SHORT ANSWER

A student writes:
"$\angle A = \angle P$ (given). $\angle B = \angle Q$ (given). So $\triangle ABC \sim \triangle PQR$ (AA). Therefore $AB = PQ$ and $BC = QR$."
(a) Which parts are correct?
(b) Which part is wrong, and what should it say?
(c) Under what single additional condition would the student's final line actually be true?

Write your working in your book.

Stretch Challenge · The altitude to the hypotenuse

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CHALLENGE

$\triangle ABC$ has a right angle at $A$. The point $D$ lies on $BC$ so that $AD$ is perpendicular to $BC$.
(a) Prove that $\triangle ABD \sim \triangle CBA$.
(b) Hence show that $AB^2 = BD \times BC$.
(c) By proving a second similarity, obtain a matching result for $AC$, and add the two results together. What theorem have you proved?

Quick Review

recap

Three tests

AA, SSS in ratio, SAS in ratio

AA is enough

The third angle follows from the angle sum

Conclusions

Ratios of matching sides, never equalities

Congruence

Similarity with scale factor $1$

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