If bisectors of opposite angles of a cyclic quadrilateral $ABCD$ intersect the circle circumscribing it at the points $P$ and $Q$,prove that $PQ$ is a diameter of the circle.

(N/A) Let the bisectors of opposite angles $\angle A$ and $\angle C$ of a cyclic quadrilateral $ABCD$ intersect the circle at points $P$ and $Q$ respectively.
We need to prove that $PQ$ is a diameter of the circle.
Join $AQ$ and $DQ$.
Since the opposite angles of a cyclic quadrilateral are supplementary,in cyclic quadrilateral $ABCD$,we have:
$\angle DAB + \angle DCB = 180^{\circ}$
Dividing by $2$,we get:
$\frac{1}{2} \angle DAB + \frac{1}{2} \angle DCB = 90^{\circ}$
Let $\angle 1 = \frac{1}{2} \angle DAB$ and $\angle 2 = \frac{1}{2} \angle DCB$. Thus,$\angle 1 + \angle 2 = 90^{\circ}$.
Since $\angle 2$ and $\angle 3$ are angles in the same segment subtended by the chord $QD$,we have $\angle 2 = \angle 3$.
Substituting this into the equation,we get $\angle 1 + \angle 3 = 90^{\circ}$.
This implies $\angle PAQ = 90^{\circ}$.
Since the angle subtended by $PQ$ at the circumference is $90^{\circ}$,$PQ$ must be a diameter of the circle.