$\textrm{a) Xét ΔABD và ΔACD, có}$
\(\left[ \begin{array}{l}AB = AC (gt)\\BD = CD (gt)\\\textrm{AD cạnh chung}\end{array} \right.\)
$⇒ΔABD = ΔACD (c.c.c)$
$\textrm{b) Vì ΔABD = ΔACD (theo a)}$
$→ \widehat{O1} = \widehat{O2}$ $\textrm{(2 góc tương ứng)}$
$\textrm{Mà}$ $\widehat{O1} + \widehat{O2} = 180^{0}$
$→ \widehat{O1} = \widehat{O2} = \frac{180^{0}}{2} = 90^{0}$
$→ AD ⊥ BC$