Similar Triangles

Two triangles are similar when they have exactly the same shape, even if one is bigger than the other, like a photo and its enlargement. Similar triangles let you find lengths you can’t measure directly, such as the height of a tree or the width of a river. They are also the idea that the whole of trigonometry is built on.

Key ideas

Congruent vs similar

• Congruent triangles have the same shape and the same size. All corresponding angles are equal and all corresponding sides are equal. You could place one exactly on top of the other.
• Similar triangles have the same shape but not necessarily the same size. All corresponding angles are equal, and corresponding sides are in the same ratio.

Every pair of congruent triangles is also similar (with a scale factor of $1$), but similar triangles are usually not congruent.

Properties of similar triangles

If $\triangle ABC$ is similar to $\triangle DEF$, written $\triangle ABC \sim \triangle DEF$, then:

• corresponding angles are equal: $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$
• corresponding sides are proportional:

$$\frac{DE}{AB} = \frac{EF}{BC} = \frac{DF}{AC} = k$$

The common ratio $k$ is the scale factor. If $k \gt 1$, $\triangle DEF$ is an enlargement of $\triangle ABC$; if $0 \lt k \lt 1$, it is a reduction.

$\triangle ABC \sim \triangle DEF$: matching arcs mark equal angles, and every side of $\triangle DEF$ is $2.5$ times as long.

The order of the letters matters

A similarity statement tells you which parts match. In $\triangle ABC \sim \triangle DEF$, the first letters ($A$ and $D$) match, the second letters ($B$ and $E$) match, and the third letters ($C$ and $F$) match. So side $AB$ (letters 1 and 2) matches side $DE$ (letters 1 and 2), and so on. Always write the vertices in matching order.

How to tell that two triangles are similar

You don’t need to check all six facts. Any one of these is enough:

Condition	What you check
AA (angle–angle)	Two pairs of corresponding angles are equal. (The third pair must then be equal too, since the angles in a triangle add to $180^\circ$.)
SSS (three proportional sides)	All three pairs of corresponding sides have the same ratio.
SAS (two proportional sides and the angle between them)	Two pairs of sides have the same ratio, and the angles between them are equal.

AA is the one you’ll use most. Look for right angles, shared angles, vertically opposite angles, and angles made by parallel lines.

Solving with proportions

To find a missing side:

1. Make sure the triangles are similar, and match up the vertices.
2. Find the scale factor from a pair of corresponding sides you know, or write a proportion.
3. Solve for the unknown, then check that the answer is sensible (bigger triangle, bigger sides).

Worked examples

Example 1: Using a similarity statement

In the figure above, $\triangle ABC \sim \triangle DEF$, with $AB = 4$ cm, $BC = 6$ cm, $AC = 5$ cm, $DE = 10$ cm and $\angle B = 56^\circ$. Find $\angle E$, $EF$ and $DF$.

Solution. $B$ and $E$ are both second letters, so they are corresponding angles: $\angle E = \angle B = 56^\circ$.

$AB$ and $DE$ correspond, so the scale factor from $\triangle ABC$ to $\triangle DEF$ is

$$k = \frac{DE}{AB} = \frac{10}{4} = 2.5$$

Multiply each side of $\triangle ABC$ by $2.5$:

$$EF = 2.5 \times BC = 2.5(6) = 15 \text{ cm}, \qquad DF = 2.5 \times AC = 2.5(5) = 12.5 \text{ cm}$$

Check: $\dfrac{15}{6} = 2.5$ and $\dfrac{12.5}{5} = 2.5$. ✓

Example 2: Are these triangles similar?

Decide whether each pair of triangles is similar.

• (a) A triangle with sides $6$, $8$, $12$ and a triangle with sides $9$, $12$, $18$.
• (b) A triangle with sides $5$, $7$, $9$ and a triangle with sides $10$, $14$, $20$.

Solution. Match the shortest with the shortest, the middle with the middle, and the longest with the longest, then compare the ratios.

(a)

$$\frac{9}{6} = 1.5, \qquad \frac{12}{8} = 1.5, \qquad \frac{18}{12} = 1.5$$

All three ratios are equal, so the triangles are similar (SSS), with scale factor $1.5$.

(b)

$$\frac{10}{5} = 2, \qquad \frac{14}{7} = 2, \qquad \frac{20}{9} \approx 2.22$$

The third ratio is different, so the triangles are not similar. Two matching ratios aren’t enough: all three must agree.

Example 3: A line parallel to one side

In $\triangle ABC$, point $D$ is on $AB$ and point $E$ is on $AC$, with $DE$ parallel to $BC$. If $AD = 4$, $DB = 6$, $DE = 5$ and $AE = 3$, find $BC$ and $EC$.

Solution. First show the triangles are similar. $\triangle ADE$ and $\triangle ABC$ share $\angle A$. Because $DE \parallel BC$, the corresponding angles $\angle ADE$ and $\angle ABC$ are equal. Two pairs of equal angles means

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

The side of the big triangle is the whole side: $AB = AD + DB = 4 + 6 = 10$. So the scale factor from the small triangle to the big one is

$$k = \frac{AB}{AD} = \frac{10}{4} = 2.5$$

Then

$$BC = 2.5 \times DE = 2.5(5) = 12.5, \qquad AC = 2.5 \times AE = 2.5(3) = 7.5$$

and $EC = AC - AE = 7.5 - 3 = 4.5$.

Check: $\dfrac{AE}{AD} = \dfrac{3}{4}$ and $\dfrac{AC}{AB} = \dfrac{7.5}{10} = \dfrac{3}{4}$. ✓

Example 4: Measuring a tree with a metre stick

On a sunny afternoon, a metre stick held upright casts a shadow $1.6$ m long. At the same moment, a tree casts a shadow $14.4$ m long. How tall is the tree?

Solution. The stick and the tree both stand straight up, so each makes a right angle with the ground. The sun’s rays arrive at the same angle for both, because the sun is so far away. Two pairs of equal angles, so the two triangles are similar (AA).

Same angle of sunlight, same right angle: the two triangles are similar. (Not to scale.)

Match height with height and shadow with shadow:

$$\begin{aligned} \frac{h}{1.0} &= \frac{14.4}{1.6} \\ h &= 1.0 \times 9 \\ h &= 9.0 \end{aligned}$$

The tree is about $9.0$ m tall.

Check: the tree’s shadow is $9$ times as long as the stick’s, so the tree should be $9$ times as tall as the stick. ✓

Common mistakes

Matching the wrong vertices. If you write $\triangle ABC \sim \triangle DEF$ but $\angle A$ actually equals $\angle E$, every proportion you write will be wrong. Find the equal angles first, then write the statement so that equal angles are in the same positions.

Using part of a side instead of the whole side. In Example 3, the side of the big triangle is $AB = 10$, not $DB = 6$. When one triangle sits inside another, redraw the two triangles separately so you can see each one’s full sides.

Flipping one ratio but not the other. In a proportion, both fractions must compare the same triangles in the same order, for example $\dfrac{\text{big}}{\text{small}} = \dfrac{\text{big}}{\text{small}}$. Writing $\dfrac{\text{big}}{\text{small}} = \dfrac{\text{small}}{\text{big}}$ gives a wrong answer.

Thinking one equal angle is enough. Two triangles that share just one angle can have very different shapes. You need two pairs of equal angles (AA), or the side conditions.

Checking only two side ratios. For SSS similarity, all three ratios must be equal. Example 2(b) shows two ratios agreeing while the third doesn’t.

Mixing units. If one length is in centimetres and the other is in metres, convert before you write the proportion.

Practice

1. (Warm-up) True or false? Explain.

• (a) Any two congruent triangles are similar.
• (b) Any two similar triangles are congruent.
• (c) Any two equilateral triangles are similar.

Solution

(a) True. Congruent triangles have equal angles and equal sides, so their sides are in the ratio $1 : 1$. They are similar with scale factor $1$.

(b) False. Similar triangles can be different sizes. A triangle with sides $3$, $4$, $5$ is similar to one with sides $6$, $8$, $10$, but they are not congruent.

(c) True. Every angle in an equilateral triangle is $60^\circ$, so any two equilateral triangles have equal angles (AA).

2. (Warm-up) $\triangle PQR \sim \triangle XYZ$. List the three pairs of equal angles, and write the proportion that links the sides.

Solution

Match the letters by position: $\angle P = \angle X$, $\angle Q = \angle Y$, $\angle R = \angle Z$.

$$\frac{XY}{PQ} = \frac{YZ}{QR} = \frac{XZ}{PR}$$

3. (Warm-up) $\triangle ABC \sim \triangle DEF$ with $AB = 3$ cm, $BC = 4$ cm and $DE = 7.5$ cm. Find the scale factor and $EF$.

Solution

$$k = \frac{DE}{AB} = \frac{7.5}{3} = 2.5, \qquad EF = 2.5 \times BC = 2.5(4) = 10 \text{ cm}$$

4. (Core) Is a triangle with sides $8$, $10$, $14$ similar to a triangle with sides $12$, $15$, $20$? Explain.

Solution

Match shortest to shortest, middle to middle, longest to longest:

$$\frac{12}{8} = 1.5, \qquad \frac{15}{10} = 1.5, \qquad \frac{20}{14} \approx 1.43$$

The ratios are not all equal, so the triangles are not similar. (For them to be similar, the longest side would have to be $1.5 \times 14 = 21$.)

5. (Core) In $\triangle ABC$, $\angle A = 40^\circ$ and $\angle B = 75^\circ$. In $\triangle PQR$, $\angle P = 40^\circ$ and $\angle R = 65^\circ$.

• (a) Show that the triangles are similar, and write a correct similarity statement.
• (b) If $AB = 6$, $BC = 8$ and $PQ = 9$, find $QR$.

Solution

(a) Find the third angle in each triangle:

$$\angle C = 180^\circ - 40^\circ - 75^\circ = 65^\circ, \qquad \angle Q = 180^\circ - 40^\circ - 65^\circ = 75^\circ$$

So $\angle A = \angle P = 40^\circ$, $\angle B = \angle Q = 75^\circ$ and $\angle C = \angle R = 65^\circ$. By AA, $\triangle ABC \sim \triangle PQR$ (in that order, so equal angles line up).

(b) $AB$ corresponds to $PQ$, so $k = \dfrac{PQ}{AB} = \dfrac{9}{6} = 1.5$. $BC$ corresponds to $QR$:

$$QR = 1.5 \times 8 = 12$$

6. (Core) To find the height of her school, Amira lays a small mirror flat on the ground $12$ m from the base of the wall. She steps back until she can see the top of the wall in the mirror. At that point she is $1.5$ m from the mirror, and her eyes are $1.6$ m above the ground. Light reflects off a mirror at equal angles, so the two triangles are similar. How tall is the wall?

Solution

Amira and the wall both make right angles with the ground, and the angles the light makes with the mirror are equal. So the triangle formed by Amira, her eyes and the mirror is similar to the triangle formed by the wall, its top and the mirror (AA).

Match height with height and distance from the mirror with distance from the mirror:

$$\begin{aligned} \frac{h}{1.6} &= \frac{12}{1.5} \\ h &= 1.6 \times 8 \\ h &= 12.8 \end{aligned}$$

The wall is about $12.8$ m tall.

7. (Core) A scale model of a triangular roof truss has sides $18$ cm, $24$ cm and $30$ cm. On the real roof, the longest side is $7.5$ m. Find the lengths of the other two sides of the real truss.

Solution

Use the same units first: $30$ cm $= 0.30$ m. The scale factor from the model to the real truss is

$$k = \frac{7.5}{0.30} = 25$$$$24 \text{ cm} \times 25 = 600 \text{ cm} = 6.0 \text{ m}, \qquad 18 \text{ cm} \times 25 = 450 \text{ cm} = 4.5 \text{ m}$$

The other sides are $6.0$ m and $4.5$ m. Check: $\dfrac{6.0}{0.24} = 25$ and $\dfrac{4.5}{0.18} = 25$. ✓

8. (Challenge) To measure the width of a river, a surveyor picks a tree $T$ on the far bank. On the near bank she marks $P$ directly across from $T$, so that $TP$ is perpendicular to the bank. She walks $30$ m along the bank to $Q$ and puts in a stake, then $10$ m further to $R$. From $R$ she walks straight away from the river, perpendicular to the bank, until she reaches the point $S$ where the stake at $Q$ lines up exactly with the tree. $RS = 12$ m. How wide is the river?

Solution

$\angle TPQ = \angle SRQ = 90^\circ$, and $\angle TQP = \angle SQR$ because they are vertically opposite angles (the line $TS$ crosses the bank at $Q$). By AA, $\triangle TPQ \sim \triangle SRQ$.

Match the sides by the letter order: $TP$ matches $SR$, and $PQ$ matches $RQ$.

$$\begin{aligned} \frac{TP}{SR} &= \frac{PQ}{RQ} \\ \frac{TP}{12} &= \frac{30}{10} \\ TP &= 12 \times 3 = 36 \end{aligned}$$

The river is about $36$ m wide.

9. (Challenge) In $\triangle ABC$, $D$ is on $AB$ and $E$ is on $AC$, with $DE \parallel BC$. If $AD = x$, $DB = 6$, $AE = 4$ and $EC = 8$, find $x$.

Solution

As in Example 3, $\triangle ADE \sim \triangle ABC$ (AA). Use whole sides for the big triangle: $AB = x + 6$ and $AC = 4 + 8 = 12$.

$$\begin{aligned} \frac{AD}{AB} &= \frac{AE}{AC} \\ \frac{x}{x + 6} &= \frac{4}{12} \\ 12x &= 4(x + 6) && \text{cross-multiply} \\ 12x &= 4x + 24 \\ 8x &= 24 \\ x &= 3 \end{aligned}$$

Check: $\dfrac{3}{9} = \dfrac{1}{3}$ and $\dfrac{4}{12} = \dfrac{1}{3}$. ✓