Prove that in a parallelogram,the bisectors of any two consecutive angles intersect at right angles.

(N/A) Given : $A$ parallelogram $ABCD$ such that the bisectors of consecutive angles $A$ and $B$ intersect at $P$.
To Prove : $\angle APB = 90^{\circ}$.
Proof : Since $ABCD$ is a parallelogram,$AD \parallel BC$.
Now,$AD \parallel BC$ and transversal $AB$ intersects them. Therefore,$\angle A + \angle B = 180^{\circ}$ (Sum of consecutive interior angles is $180^{\circ}$).
Dividing by $2$,we get $\frac{1}{2} \angle A + \frac{1}{2} \angle B = 90^{\circ}$.
Since $AP$ is the bisector of $\angle A$ and $BP$ is the bisector of $\angle B$,we have $\angle PAB + \angle PBA = 90^{\circ}$.
In $\Delta APB$,the sum of angles is $180^{\circ}$:
$\angle PAB + \angle APB + \angle PBA = 180^{\circ}$.
Substituting the value,$90^{\circ} + \angle APB = 180^{\circ}$.
Therefore,$\angle APB = 180^{\circ} - 90^{\circ} = 90^{\circ}$.