Similar triangles

Con­gru­ent fig­ures have the same shape and the same size. Sim­i­lar fig­ures have the same shape but pos­si­bly dif­fer­ent sizes: a pho­to­graph and its enlarge­ment, a map and the land it shows, a model aero­plane and the real one. Sim­i­lar tri­an­gles let us mea­sure things we can­not reach, such as the height of a tower from its shadow. This les­son proves the basic pro­por­tion­al­ity the­o­rem and sets out the three tests for sim­i­lar tri­an­gles.

What sim­i­lar means#

Two tri­an­gles are sim­i­lar if

1. their cor­re­spond­ing angles are equal, and
2. their cor­re­spond­ing sides are in the same ratio (pro­por­tional).

We write $\triangle ABC \sim \triangle PQR$, and as with con­gru­ence the order of let­ters gives the cor­re­spon­dence. If $\triangle ABC \sim \triangle PQR$, then

$$\displaystyle \angle A = \angle P, \quad \angle B = \angle Q, \quad \angle C = \angle R, \qquad \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$$

Every pair of con­gru­ent tri­an­gles is sim­i­lar (with ratio $1$), but sim­i­lar tri­an­gles need not be con­gru­ent.

The basic pro­por­tion­al­ity the­o­rem#

The­o­rem (Thales). If a line is drawn par­al­lel to one side of a tri­an­gle, cut­ting the other two sides at dis­tinct points, it divides those two sides in the same ratio.

In tri­an­gle $ABC$, if $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$, then $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$.

Proof. Join $BE$ and $CD$. We com­pare areas of tri­an­gles, using the fact that tri­an­gles with the same height have areas in the ratio of their bases.

Tri­an­gles $ADE$ and $BDE$ have the same height from $E$ to line $AB$, so

$$\displaystyle \frac{\text{ar}(ADE)}{\text{ar}(BDE)} = \frac{AD}{DB}$$

Tri­an­gles $ADE$ and $CDE$ have the same height from $D$ to line $AC$, so

$$\displaystyle \frac{\text{ar}(ADE)}{\text{ar}(CDE)} = \frac{AE}{EC}$$

Tri­an­gles $BDE$ and $CDE$ stand on the same base $DE$ and lie between the same par­al­lels $DE$ and $BC$, so they have equal areas. The two frac­tions there­fore have the same denom­i­na­tor, and $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$.

Con­verse. If a line divides two sides of a tri­an­gle in the same ratio, it is par­al­lel to the third side.

Exam­ple 1#

In tri­an­gle $ABC$, $DE \parallel BC$, $AD = 3$ cm, $DB = 2$ cm and $AE = 4.5$ cm. Find $EC$.

$$\displaystyle \frac{AD}{DB} = \frac{AE}{EC} \implies \frac{3}{2} = \frac{4.5}{EC} \implies EC = \frac{2 \times 4.5}{3} = 3 \text{ cm}$$

Exam­ple 2#

In tri­an­gle $ABC$, $DE \parallel BC$, with $AD = x + 1$, $DB = x - 1$, $AE = x + 4$ and $EC = x$. Find $x$.

$$\displaystyle \begin{aligned} \frac{x + 1}{x - 1} &= \frac{x + 4}{x} \\ x(x + 1) &= (x - 1)(x + 4) \\ x^2 + x &= x^2 + 3x - 4 \\ 4 &= 2x \\ x &= 2 \end{aligned}$$

Then $AD = 3$, $DB = 1$, $AE = 6$, $EC = 2$, and indeed $\displaystyle \frac{3}{1} = \frac{6}{2}$.

Exam­ple 3#

$D$ and $E$ are points on sides $AB$ and $AC$ of tri­an­gle $ABC$, with $AD = 4$ cm, $DB = 6$ cm, $AE = 5$ cm and $EC = 7.5$ cm. Is $DE$ par­al­lel to $BC$?

$$\displaystyle \frac{AD}{DB} = \frac{4}{6} = \frac{2}{3}, \qquad \frac{AE}{EC} = \frac{5}{7.5} = \frac{2}{3}$$

The ratios are equal, so by the con­verse, $DE \parallel BC$.

Tests for sim­i­lar tri­an­gles#

We do not need to check all angles and all ratios. Any one of these is enough:

In Exam­ple 1, tri­an­gles $ADE$ and $ABC$ are sim­i­lar by AA: they share $\angle A$, and $\angle ADE = \angle ABC$ (cor­re­spond­ing angles, $DE \parallel BC$).

Exam­ple 4#

A ver­ti­cal pole $6$ m high casts a shadow $4$ m long on level ground. At the same time a tower casts a shadow $28$ m long. Find the height of the tower.

The sun's rays make the same angle with the ground for both, and both stand at right angles to the ground. So the tri­an­gle formed by the pole, its shadow and the ray is sim­i­lar to the one formed by the tower (AA). Cor­re­spond­ing sides are pro­por­tional:

$$\displaystyle \frac{\text{height of tower}}{6} = \frac{28}{4} \implies \text{height} = 6 \times 7 = 42 \text{ m}$$

Exam­ple 5#

Are tri­an­gles with sides $4$, $6$, $8$ cm and $6$, $9$, $12$ cm sim­i­lar?

Match the short­est with the short­est, and so on:

$$\displaystyle \frac{6}{4} = 1.5, \qquad \frac{9}{6} = 1.5, \qquad \frac{12}{8} = 1.5$$

All three ratios are equal, so the tri­an­gles are sim­i­lar by SSS.

Exam­ple 6#

In tri­an­gle $ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD = 2$ cm, $AB = 5$ cm and $BC = 15$ cm, find $DE$.

$\triangle ADE \sim \triangle ABC$ (AA, as above), so

$$\displaystyle \frac{DE}{BC} = \frac{AD}{AB} \implies \frac{DE}{15} = \frac{2}{5} \implies DE = 6 \text{ cm}$$

Note the ratio used: $\displaystyle \frac{AD}{AB}$ com­pares a side of the small tri­an­gle with the match­ing side of the large one. The ratio $\displaystyle \frac{AD}{DB}$ from Thales' the­o­rem would be wrong here.

Prac­tice#

1. In tri­an­gle $ABC$, $DE \parallel BC$, $AD = 4$ cm, $DB = 6$ cm and $AE = 6$ cm. Find $EC$.
2. In tri­an­gle $ABC$, $DE \parallel BC$, $AD = 2.4$ cm, $AE = 3.2$ cm and $EC = 4.8$ cm. Find $DB$.
3. $D$ and $E$ lie on $AB$ and $AC$ with $AD = 5$, $DB = 8$, $AE = 7.5$ and $EC = 12$ (in cm). Is $DE \parallel BC$?
4. $D$ and $E$ lie on $AB$ and $AC$ with $AD = 3$, $DB = 5$, $AE = 4$ and $EC = 6$ (in cm). Is $DE \parallel BC$?
5. In tri­an­gle $ABC$, $DE \parallel BC$, with $AD = x + 2$, $DB = x$, $AE = x + 6$ and $EC = x + 2$. Find $x$.
6. A stick $1.5$ m long, held upright, casts a shadow $2$ m long. At the same time a build­ing casts a shadow $24$ m long. How tall is the build­ing?
7. Are tri­an­gles with sides $3$, $5$, $6$ cm and $4.5$, $7.5$, $9$ cm sim­i­lar? Give the rea­son.
8. $\triangle ABC \sim \triangle DEF$, with $AB = 4$ cm, $BC = 5$ cm, $CA = 6$ cm and $DE = 10$ cm. Find $EF$ and $FD$.
9. In tri­an­gle $ABC$, $DE \parallel BC$ with $D$ on $AB$, $AD = 3$ cm, $AB = 9$ cm and $BC = 12$ cm. Find $DE$.

Answers#