Question

Question 5
ABC is an isosceles triangle with AB = AC. Drawn AP $\perp$ BC to show that $\angle B=\angle C$.

Answer 1

In $\Delta$APC and $\Delta$APB
AB = AC (Given)
AP = AP (Common)
$\angle APB=\angle APC$ (Each ${90}^{\circ }$)
$\therefore \Delta APC\cong \Delta APB$ (RHS congruence)
$\Rightarrow \angle B=\angle C$ (By CPCT)