Question
The given figure shows a triangle $A B C$ in which $A D$ is perpendicular to side $B C$ and $B D=C D$. Prove that:
(i) $\triangle A B D \cong \triangle A C D$
(ii) $A B=A C$
(iii) $\angle B=\angle C$

(i) $\triangle A B D \cong \triangle A C D$
(ii) $A B=A C$
(iii) $\angle B=\angle C$

