# Similarity of Triangles - Theorems and Proofs

Triangle is a polygon which has three sides and three vertices. Triangles having same shape and size are said to be congruent. Similarity of triangles uses the concept of similar shape and finds great applications. Triangles are said to be similar if:

a. Their corresponding angles are equal.

b. Their corresponding sides are in the same ratio.

Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is said to be parallel to the third side.

Let ABC be a triangle with sides AB, AC and BC. DE divides any two sides of the triangle in the same ratio.

Given: Line divides a triangle in the same ratio. Thus,

$\frac{AD}{DB}$ = $\frac{AE}{EC}$

Proof:

$\frac{AD}{DB}$ = $\frac{AE}{EC}$ (Given)—-(1)

Let us assume that DE is not parallel to BC. Now we draw DE’ which is assumed to be parallel to BC. So,

$\frac{AD}{DB}$ = $\frac{AE’}{E’C}$ (Property of similar triangles)—-(2)

Therefore, from (1) and (2)

$\frac{AE}{EC}$ = $\frac{AE’}{E’C}$

Now we add 1 to both sides,

$\frac{AE}{EC}~+~1$ = $\frac{AE’}{E’C}~ + ~1$

$\Rightarrow~\frac{AE}{EC}~+~\frac{EC}{EC}$ = $\frac{AE’}{E’C}~+~\frac{E’C}{E’C}$

$\Rightarrow~\frac{AE~+~EC}{EC}$ = $\frac{AE’ + E’C}{E’C}$

According to the figure $AE + EC$ = $AC$ and $AE’ + E’C$ = $AC$, substituting these values in the equation above:

$\frac{AC}{EC}$ = $\frac{AC}{E’C}$

This directly implies that $EC$ = $E’C$ and $E$ = $E’$ meaning that they are the same point.

Hence DE is parallel to BC. This proves the similarity of triangles.

Let us take an example to observe the property of similarity of triangles:

Illustration 1:PQRS is a trapezium with PQ parallel to RS. The point X and Y are on the non-parallel sides PS and QR respectively such that XY is parallel to PQ. Show that $\frac{PX}{XS}$ = $\frac{QY}{YR}$.

Solution: Let us first join PR in order to intersect XY at Z.

$PQ$ || $RS$ and $XY$ || $PQ$ (given)

So, $XY$ || $RS$ (lines parallel to same line are parallel to each other)

In $\triangle~PSR$,

$XZ$ || $SR$ (as $XY$ || $SR$)

So, $\frac{PX}{XS}$ = $\frac{PZ}{ZR}$ …….. (3)

Similarly, from $\triangle~PRQ$

$\frac{RZ}{ZP}$ = $\frac{RY}{YQ}$

$\frac{PZ}{ZR}$ = $\frac{QY}{YR}$ …………. (4)

From equation (3) and (4),

$\frac{PX}{XS}$ = $\frac{QY}{YR}$