Circle Geometry Assistance

Question

Two circles $C_1$ and $C_2$ meet at the points $P$ and $Q$.

A circle $C_3$, with centre at point $P$, meets $C_1$ and $C_2$ at points $A$, $B$, $C$ and $D$ respectively.

Prove that $\angle AQD= \angle BQC$.

Answer 1

Using the inscribed angle theorem,

$$\begin{align}\angle{AQD}&=\angle{AQP}+\angle{DQP}\\&=\angle{ABP}+\angle{DCP}\\&=\angle{BAP}+\angle{CDP}\qquad(\leftarrow \text{$\triangle{PAB},\triangle{PCD}$ are isosceles triangles})\\&=\left(180^\circ-\angle{BQP}\right)+\left(180^\circ-\angle{CQP}\right)\\&=360^\circ-\angle{BQP}-\angle{CQP}\\&=\angle{BQC}\end{align}$$