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$ \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,
$\angle B=\angle C$
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