When we plan to construct a building, we make a plan and draw a diagram which we call as the plan of the house or the building. This will help us to understand the area required, volume of the room, amount materials required etc. The diagram we obtain is similar to the original which is diminished in size. In the case of similar figures, the ratio of the corresponding sides remains the same. We use the same property in similar triangles.

In this section we shall discuss with the various properties used to verify whether the triangles are similar. We shall also discuss with some of the proving questions and practice problems.

## Definition of Similar Triangles

Definition of Similar Triangles: Two triangles are said to be similar if their corresponding angles are equal and the corresponding sides [angles opposite to equal angles] are proportional.

### Properties of Similar Triangles

1. When two triangles are similar, their corresponding angles are equal.
In the above two triangles,
$\angle A$ = $\angle P$
$\angle B$ = $\angle Q$
$\angle C$ = $\angle R$

2. When two triangles are similar, their corresponding sides are proportional.

$\frac{BC}{QR}$ = $\frac{AC}{PR}$ = $\frac{AB}{PQ}$

### Similar Triangles Formula

1. If two triangles are similar then, the ratio of the sides of the two similar triangles is equal to the ratio of their corresponding altitudes.

In the above figure $\Delta ABC$ $\sim$ $\Delta PQR$,

AD and PS are the altitudes to the opposite sides.

Therefore, we have, $\frac{AB}{PQ}$ = $\frac{BC}{QR}$ = $\frac{AC}{PR}$ = $\frac{AD}{PS}$

2. If two triangles are similar then, the ratio of the sides of the two similar triangles is equal to the ratio of their corresponding medians.

In the above figure $\Delta ABC$ $\sim$ $\Delta PQR$,

AD and PS are the medians to the opposite sides.

Therefore, we have, $\frac{AB}{PQ}$ = $\frac{BC}{QR}$ = $\frac{AC}{PR}$ = $\frac{AD}{PS}$

3. If two triangles are similar then, the ratio of the sides of the two similar triangles is equal to the ratio of their corresponding angular bisectors.

In the above figure $\Delta ABC$ $\sim$ $\Delta PQR$,

AD and PS are the bisectors of the angles $\angle A$ and $\angle P$ respectively.

Therefore, we have, $\frac{AB}{PQ}$ = $\frac{BC}{QR}$ = $\frac{AC}{PR}$ = $\frac{AD}{PS}$

4. Ratio of the areas of the two similar triangles is equal to the ratio of the squares of the corresponding sides.
In the above triangles, $\Delta ABC$ $\sim$ $\Delta PQR$

Therefore, we have, $\frac{Area(\Delta ABC)}{Area(\Delta PQR)}$ = $\frac{AB^{2}}{PQ^{2}}$ = $\frac{BC^{2}}{QR^{2}}$ = $\frac{AC^{2}}{PR^{2}}$

### Similar Triangles Theorem

Theorem 1: (AAA Criterion) If in two triangles the corresponding angles are equal then the triangles are similar. This is also called as AAA crteria.

In $\Delta ABC$ and $\Delta KLM$,
$\angle A$ = $\angle K$
$\angle B$ = $\angle L$
$\angle C$ = $\angle M$,
Therefore, $\Delta ABC$ $\sim$ $\Delta KLM$

Theorem 2: (AA Criterion)  If in two triangles, two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Proof: When two angles of one a\triangle are equal to two angles of another triangle,
we have $\angle B$ = $\angle L$
$\angle C$ = $\angle M$
In $\Delta ABC$ and $\Delta KLM$, we have
$\angle A$ + $\angle B$ + $\angle C$ = 180o  ------------------------------ (1)
and $\angle K$ + $\angle L$ + $\angle M$ = 180o -----------------------------------(2)
Equating the two statements ( 1 ) and ( 2 ), we get,
$\angle A$ + $\angle B$ + $\angle C$  = $\angle K$ + $\angle L$ + $\angle M$
=>                           $\angle A$  = $\angle K$ [ since, $\angle C$ = $\angle M$ and $\angle B$ = $\angle L$ ]
As the three angles are one triangle are respectively equal to the three angles of another triangle, we have
$\Delta ABC$ $\sim$ $\Delta KLM$

Theorem 3: (SSS Criterion)  If the sides of one triangle are proportional to the sides of another triangle, then the triangles are said to be similar by SSS criteria.

$\Delta ABC$ and $\Delta KLM$, we have

$\frac{AB}{KL}$ = $\frac{9}{6}$ = 1.5

$\frac{BC}{LM}$ = $\frac{6}{4}$ = 1.5

$\frac{AC}{KM}$ = $\frac{12}{8}$ = 1.5

Therefore, we have $\frac{AB}{KL}$$\frac{BC}{LM}$$\frac{AC}{KM}$

As the sides of the triangle are proportional, we have $\Delta ABC$ $\sim$ $\Delta KLM$

Theorem 4: (SAS Criterion)If in two triangles, one pair of corresponding sides are proportional and included angles are equal, then the two triangles are similar.

$\Delta ABC$ and $\Delta KLM$, we have

$\frac{AB}{KL}$ = $\frac{9}{6}$ = 1.5

$\frac{BC}{LM}$ = $\frac{6}{4}$ = 1.5

and         $\angle B$ = $\angle L$

Therefore, we have $\Delta ABC$ $\sim$ $\Delta KLM$

### Right Triangles

1. If an acute angle of one triangle is equal to one of the acute angle of another triangle then the two triangles are said to the similar by AA similarity.

In $\Delta ABC$ and  $\Delta PQR$,
$\angle B$ = $\angle Q$ = 90o
$\angle C$ = $\angle R$
Therefore, $\Delta ABC$ $\sim$ $\Delta PQR$, by AA similarity Criteria.

2. If the sides including the right angle of one right  triangle are proportional to the sides including the right angle of another right triangle, then the triangles are said to be similar by SAS similarity.

In $\Delta ABC$ and  $\Delta PQR$,

$\angle B$ = $\angle Q$ = 90o

$\frac{AB}{PQ}$ = $\frac{BC}{QR}$

Therefore, $\Delta ABC$ $\sim$ $\Delta PQR$, by SAS similarity Criteria.

## Proving Triangles Similar

Similar Triangle Proofs
Prove that the line segments joining the midpoints of the sides of a triangle form four triangles, each of which is similar to the original triangle.

Given: In $\Delta ABC$, D, E and F are the midpoints of BC, CA and AB respectively.
To Prove: Each of the triangles, $\Delta AFE$, $\Delta FBD$, $\Delta EDC$ and $\Delta DEF$ are similar to $\Delta ABC$
Proof:
 Serial No. Statement Reason 1. In $\Delta ABC$, E, F are the midpoints of AB and AC respectively Given 2. EF // BC by Mid point theorem 3. $\angle AFE$ = $\angle B$ corresponding angles 4. In $\Delta AFE$ and $\Delta ABC$, 5. $\angle A$ = $\angle A$ Common 6. $\angle AFE$ = $\angle B$ from ( 3 ) 7. Therefore, $\Delta AFE$ $\sim$ $\Delta ABC$ by AA similarity  Criteria 8. Similarly, $\Delta FBD$ $\sim$ $\Delta ABC$               $\Delta EDC$ $\sim$ $\Delta ABC$ 9. AFDE is a parallelogram AF // ED and AE // FD. 10. $\angle A$ = $\angle EDF$ opposite angles of a parallelogram are equal 11. Similarly, $\angle B$ = $\angle FED$ 12. $\Delta ABC$  $\sim$  $\Delta DEF$ by AA Similarity Criteria

Therefore, each of the triangles, $\Delta AFE$, $\Delta FBD$, $\Delta EDC$ and $\Delta DEF$ are similar to $\Delta ABC$

## Similar Triangles Problems

1. In the following figure, if AD $\perp$ BC and $\frac{BD}{DA}$ = $\frac{DA}{DC}$, prove that $\Delta ABC$ is a right triangle.

Given: In $\Delta ABC$, AD $\perp$  BC and $\frac{BD}{DA}$ = $\frac{DA}{DC}$.

To Prove: $\Delta ABC$ is a right triangle.
Proof :
 Serial No. Statement Reason 1. In $\Delta BDA$ and $\Delta ADC$,     $\frac{BD}{DA}$ = $\frac{DA}{DC}$ Given 2. $\angle BDA$ = $\angle ADC$ = 90o Given 3. $\Delta BDA$   $\sim$  $\Delta ADC$ By AA similarity Criteria 4. $\angle ABD$ = $\angle CAD$   $\angle ACD$ = $\angle BAD$ Corresponding angles of the similar triangles 5. $\angle ABD$ + $\angle ACD$ = $\angle CAD$ + $\angle BAD$ Adding the two angles in ( 4 ) 6. $\angle B$ + $\angle C$ = $\angle A$ 7. $\angle A$ + $\angle B$ + $\angle C$ = $\angle A$ + $\angle A$ Adding $\angle A$ on both sides 8. 180o = 2 $\angle A$ angle sum property of a triangle 9. $\angle A$ = $\frac{180}{2}$ = 90o Dividing both sides by 2 10. $\Delta ABC$ is a right triangle. Property of right angled triangle

2. In a triangle ABC, DE // BC, such that AE = $\frac{1}{4}$ AC. IF AB = 6 cm find AD.

Given: In $\Delta ABC$, DE // BC, AE = $\frac{1}{4}$ AC and  AB = 6 cm.
To Prove: $\Delta ADE$ $\sim$ $\Delta ABC$ and find AD.
Proof:
 Serial No. Statement Reason 1. In In $\Delta ADE$ and $\Delta ABC$,$\angle ADE$ = $\angle B$ DE // BC , Corresponding Angles are equal. 2. $\angle A$ = $\angle A$ common angle 3. $\Delta ADE$   $\sim$   $\Delta ABC$ by AA similarity Criteria 4. $\frac{AD}{AB}$ = $\frac{AE}{AC}$ sides of the similar triangles are proportional 5. $\frac{AD}{AB}$ = $\frac{\frac{1}{4}AC}{AC}$ Substituting AE = $\frac{1}{4}$ AC 6. $\frac{AD}{AB}$ = $\frac{1}{4}$ dividing the Numerator and the Denominator by AC. 7. AD = $\frac{1}{4}$ AB Multiplying both sides by AB 8. AD = $\frac{1}{4}$ ( 6 ) = $\frac{3}{2}$ = 1.5 Substituting AB = 6 cm

Therefore, we have AD = 1.5 cm