$ \mathrm{ABC} $ is an isosceles triangle with $ \mathrm{AB}=\mathrm{AC} $. Draw $ \mathrm{AP} \perp \mathrm{BC} $ to show that $ \angle \mathrm{B}=\angle \mathrm{C} $.

Given:

$ABC$ is an isosceles triangle with $AB=AC$.

To do:

We have to draw $AP \perp BC$ to show that $\angle B=\angle C$.

Solution:

Let us consider $\triangle ABP$ and $\triangle ACP$

We know that according to the RHS rule if the hypotenuse and one side of a right-angled triangle are equal to the corresponding hypotenuse and one side of another right-angled triangle; then both the right-angled triangle are said to be congruent.

We have,

$AB=AC$ and

As $AP \perp BC$ i.e., $AP$ is the altitude.

We get,

$\angle APB=\angle ACP=90^o$

Since, $AP$ is the common side we get,

$AP=PA$

Therefore,

$\triangle ABP \cong \triangle ACP$

We also know that,

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

This implies,

$\angle B=\angle C$