$ABC$ is a right-angled triangle with $AB = AC$. The bisector of $\angle A$ meets $BC$ at $D$. Prove that $BC = 2 AD$.

(N/A) Given: $A$ right-angled triangle $ABC$ where $AB = AC$ and the bisector of $\angle A$ meets $BC$ at $D$.
To prove: $BC = 2 AD$.
Proof: In $\triangle ABC$,since $AB = AC$ and it is a right-angled triangle,$\angle BAC = 90^{\circ}$.
In $\triangle CAD$ and $\triangle BAD$:
$AC = AB$ (Given)
$\angle CAD = \angle BAD$ (Since $AD$ is the bisector of $\angle A$)
$AD = AD$ (Common side)
By $SAS$ congruence criterion,$\triangle CAD \cong \triangle BAD$.
Therefore,$CD = BD$ (by $CPCT$).
Since $AD$ is the angle bisector of the right angle in an isosceles right triangle,$AD$ is also the median to the hypotenuse.
In a right-angled triangle,the median to the hypotenuse is half the length of the hypotenuse.
Thus,$AD = \frac{1}{2} BC$.
Multiplying both sides by $2$,we get $2 AD = BC$.
Hence,$BC = 2 AD$ is proved.