Proof With Cyclic Quadrilateral and Circle Let $ABCD$ be a cyclic quadrilateral. Let $P$ be the intersection of $\overline{AD}$ and $\overline{BC}$, and let $Q$ be the intersection of $\overline{AB}$ and $\overline{CD}$. Prove that the angle bisectors of $\angle DPC$ and $\angle AQD$ are perpendicular.

All I know about cyclic quadrilaterals is that opposite angles add up to $180^\circ.$  How can I use that to finish my proof?  Thanks in advance for answering!
 A: We need a few more letters: Let the intersection between the bisectors be at $I$, let $E, F$ be where the bisector at $P$ intersects $AB$ and $CD$ respectively, and let $G,H$ be where the bisector at $Q$ intersects $BC$ and $AD$ respectively. This gives the following figure:

(In retrospect, $E$ and $G$ were unnecessary, but hey, that's how you solve geometry puzzles: You keep adding points and lines and calculate until you find the ones you really needed.)
Let, for convinience, $\angle DAB = a$ and $\angle ADC = d$ (assuming the quadrilateral is named so that $A$ is between $D$ and $P$, and $B$ is between $A$ and $Q$). Now we chase angles.
$\angle AQD = 180^{\circ} - a-d$ and $\angle DPC = a-d$ (since $\angle DCP = 180^\circ - a$).
Since $QH$ bisects $\angle AQD$, we get $\angle DHQ = 90 + \frac a2 - \frac d2 = $ and similarily $\angle DFP = 180^\circ - \frac{a}{2} - \frac d2$. That means that in the quadrilateral $DFIH$ we know three of the angles, and they add up to
$$
\angle ADC + \angle DFP + \angle DHQ = d + 180^\circ - \frac{a}{2} - \frac d2 + 90^\circ  + \frac a{2} - \frac d2 = 270^\circ
$$
Thus the last angle in the quadrilateral must be $90^\circ$.
A: Here is another, simpler solution. Apologies for the orientation of the image.

A: Construction:- 1) Extend CP.  2) Draw PR, the external bisector of $\angle APB$.

It should be clear that $\theta = \theta’$, $\phi = \phi’$, $\omega = \omega’$, and $n = n'$ [ext. angle, cyclic quad.]
The logic:- We know that the internal and the external angle bisector are perpendicular to each other. Thus, we are done if we can prove that PR // QI. The question now is changed to “will $\omega = \alpha?$”
By exterior angle, $\omega = \dfrac {m + n’}{2} = \dfrac {(n + 2\phi) + n'}{2} = n+\phi= \alpha$.
