ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C.
Solution:

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In $\triangle \mathrm{APB}$ and $\triangle \mathrm{APC}$

$\angle \mathrm{APB}=\angle \mathrm{APC}\left(\right.$ Each $\left.90^{\circ}\right)$

$A B=A C($ Given $)$

$\mathrm{AP}=\mathrm{AP}($ Common $)$

$\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{APC}(\cup$ sing RHS congruence rule $)$

$\Rightarrow \angle B=\angle C$ (By using CPCT)
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