Let $P$ and $Q$ be adjacent corners of a parallelogram, and let those angles have measure $p$ and $q$. Let $R$ be the point at which the angle bisectors at $P$ and $Q$ meet.
In $\triangle PQR$, we have
$$180^\circ = \angle R + \angle RPQ + \angle RQP = \angle R + \frac{1}{2}p + \frac{1}{2}q = \angle R + \frac{1}{2}\left( p+q \right)$$
Adjacent angles in a parallelogram are supplementary, so $p+q=180^\circ$. Thus,
$$180^\circ = \angle R + 90^\circ \qquad \implies \qquad \angle R = 90^\circ$$
which is to say: _Adjacent angle bisectors in a parallelogram meet at right angles._