In Fig. 7.21, $ \mathrm{AC}=\mathrm{AE}, \mathrm{AB}=\mathrm{AD} $ and $ \angle \mathrm{BAD}=\angle \mathrm{EAC} $. Show that $ \mathrm{BC}=\mathrm{DE} $.
"

Given:

$AC=AE,AB=AD$ and $\angle BAD=\angle EAC$.

To do:

We have to show that $BC=DE$.

Solution:

Let us add $\angle DAC$ on both sides of $\angle BAD=\angle EAC$ we get,

$\angle BAD+\angle DAC=\angle EAC+\angle DAC$

This implies,

$\angle BAC=\angle EAD$

We know that,

According to Rule of Side-Angle-Side Congruence:

Triangles are said to be congruent if any pair of corresponding sides and their included angles are equal in both triangles.

Given,

$AC=AE$ and $AB=AD$

We also have,

$\angle BAC=\angle EAD$

Therefore,

$\angle ABC \cong \angle ADE$.

We also know,

From corresponding parts of congruent triangles: If two triangles are congruent, all of their corresponding  sides must be equal.

Therefore,

$BC=DE$.