Question

In Figure $$,\Delta F E C \cong \Delta G D B$$ and $$\angle 1=\angle 2 .$$ Prove that $$\Delta A D E \sim \Delta A B C$$

Answer 1

Since

$$\Delta F E C \cong \Delta G D B$$

$$\Rightarrow \mathrm{EC}=\mathrm{BD} \quad \quad \dots (i)$$

It is given that

$$\angle 1=\angle 2$$

$$\Rightarrow AE=AD \qquad [\text { Sides opposite to equal angles are equal] } \ldots (ii) $$

From (i) and (ii), we have

$$\dfrac{AE}{E C}=\dfrac{A D}{B D}$$

$$ \Rightarrow DE \parallel B C \quad \text { [By the converse of basic proportionality theorem] } $$

$$\Rightarrow\angle 1=\angle 3 \text { and } \angle 2=\angle 4 \quad[\text { Corresponding angles }] $$

Thus, in $$\Delta^{\prime}s\ A D E$$ and $$A B C$$, we have

$$\angle A=\angle A \quad[\text { Common }]$$

$$ \angle 1=\angle 3 $$

$$\angle 2=\angle 4 \quad {[\text { Proved above }]} $$

So, by AAA criterion of similarity, we have

$$\Delta A D E \sim \Delta A B C$$