In parallelogram $PQRS$,diagonal $PR$ bisects $\angle P$. Prove that diagonal $PR$ bisects $\angle R$ also and $PQRS$ is a rhombus.

(N/A) Given: $PQRS$ is a parallelogram where diagonal $PR$ bisects $\angle P$,i.e.,$\angle SPR = \angle QPR$.
Step $1$: Since $PQRS$ is a parallelogram,$PS \parallel QR$ and $PQ \parallel SR$.
Step $2$: Because $PS \parallel QR$ and $PR$ is a transversal,$\angle SPR = \angle PRQ$ (alternate interior angles).
Step $3$: Because $PQ \parallel SR$ and $PR$ is a transversal,$\angle QPR = \angle PRS$ (alternate interior angles).
Step $4$: Since $\angle SPR = \angle QPR$ (given),it follows that $\angle PRQ = \angle PRS$. Thus,$PR$ bisects $\angle R$.
Step $5$: In $\triangle PQR$ and $\triangle PSR$,we have $\angle QPR = \angle SPR$ and $\angle PRQ = \angle PRS$. Since $PR = PR$ (common side),the triangles are congruent by $ASA$ congruence.
Step $6$: By $CPCT$,$PQ = PS$. Since adjacent sides of a parallelogram are equal,$PQRS$ is a rhombus.