$ABCD$ is a quadrilateral such that diagonal $AC$ bisects the angles $A$ and $C.$ Prove that $AB = AD$ and $CB = CD.$

(N/A) Given: $A$ quadrilateral $ABCD$ such that $\angle 1 = \angle 2$ and $\angle 3 = \angle 4.$
To prove: $AB = AD$ and $CB = CD.$
Proof: In $\triangle ABC$ and $\triangle ADC$,we have:
$\angle 1 = \angle 2$ [Given]
$AC = AC$ [Common side]
$\angle 3 = \angle 4$ [Given]
So,by $ASA$ criterion of congruence,we have:
$\triangle ABC \cong \triangle ADC$
Therefore,$AB = AD$ [by $CPCT$]
And $CB = CD$ [by $CPCT$]
Hence,proved.