In $\Delta ABC$,$\angle A = 90^{\circ}$ and $AB = AC$. The bisector of $\angle A$ meets $BC$ at $D$. Prove that $BC = 2 AD$.

(N/A) Given: In $\Delta ABC$,$\angle A = 90^{\circ}$ and $AB = AC$. $AD$ is the bisector of $\angle A$.
Since $AB = AC$,$\Delta ABC$ is an isosceles triangle. Therefore,$\angle B = \angle C$.
In $\Delta ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.
$90^{\circ} + \angle B + \angle B = 180^{\circ} \implies 2 \angle B = 90^{\circ} \implies \angle B = 45^{\circ}$. Thus,$\angle C = 45^{\circ}$.
Since $AD$ is the angle bisector of $\angle A$,$\angle BAD = \angle CAD = 45^{\circ}$.
In $\Delta ABD$,$\angle B = 45^{\circ}$ and $\angle BAD = 45^{\circ}$. Since two angles are equal,$AD = BD$.
In $\Delta ACD$,$\angle C = 45^{\circ}$ and $\angle CAD = 45^{\circ}$. Since two angles are equal,$AD = CD$.
Since $D$ lies on $BC$,$BC = BD + CD$.
Substituting $BD = AD$ and $CD = AD$,we get $BC = AD + AD = 2 AD$.
Hence,$BC = 2 AD$ is proved.