$ABC$ is a right triangle such that $AB = AC$ and the bisector of angle $C$ intersects the side $AB$ at $D$. Prove that $AC + AD = BC$.

(N/A) Given: $A$ right triangle $ABC$ where $AB = AC$ and $CD$ is the bisector of $\angle C$.
To prove: $AC + AD = BC$.
Construction: Draw $DE \perp BC$.
Proof: In right triangle $ABC$,we have $AB = AC$ (Given).
Since $BC$ is the hypotenuse,$\angle A = 90^{\circ}$.
In $\triangle DAC$ and $\triangle DEC$:
$\angle DAC = \angle DEC = 90^{\circ}$ (Given and construction).
$\angle ACD = \angle ECD$ (Since $CD$ is the bisector of $\angle C$).
$CD = CD$ (Common side).
By $AAS$ congruence criterion,$\triangle DAC \cong \triangle DEC$.
Therefore,$AD = DE$ and $AC = CE$ (by $CPCT$).
In $\triangle ABC$,since $AB = AC$,$\angle B = \angle ACB$. Since $\angle A = 90^{\circ}$,$\angle B + \angle ACB = 90^{\circ}$,so $2\angle B = 90^{\circ}$,which means $\angle B = 45^{\circ}$.
In $\triangle DEB$,$\angle DEB = 90^{\circ}$ and $\angle B = 45^{\circ}$,so $\angle EDB = 180^{\circ} - (90^{\circ} + 45^{\circ}) = 45^{\circ}$.
Since $\angle EDB = \angle B = 45^{\circ}$,the sides opposite to them are equal: $DE = BE$.
Since $AD = DE$ and $DE = BE$,we have $AD = BE$.
Now,$BC = CE + BE$.
Substituting $CE = AC$ and $BE = AD$,we get $BC = AC + AD$.
Hence,$AC + AD = BC$ is proved.