In quadrilateral $PQRS$,$PQ = PS$ and $QR = RS$. Prove that diagonal $PR$ bisects $\angle QPS$ as well as $\angle QRS$.

(N/A) To prove that diagonal $PR$ bisects $\angle QPS$ and $\angle QRS$,we consider the two triangles $\triangle PQR$ and $\triangle PSR$.
$1$. In $\triangle PQR$ and $\triangle PSR$:
- $PQ = PS$ (Given)
- $QR = RS$ (Given)
- $PR = PR$ (Common side)
$2$. By the $SSS$ (Side-Side-Side) congruence criterion,$\triangle PQR \cong \triangle PSR$.
$3$. Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
- $\angle QPR = \angle SPR$. This implies that $PR$ bisects $\angle QPS$.
- $\angle QRP = \angle SRP$. This implies that $PR$ bisects $\angle QRS$.
Thus,it is proved that diagonal $PR$ bisects both $\angle QPS$ and $\angle QRS$.