Question 181
Circles $S$ and $T$ intersect at $P$ and $Q$, with $S$ passing through the centre of $T$. Distinct points $A$ and $B$ lie on $S$, inside $T$, and are equidistant from the centre of $T$. The line $PA$ meets $T$ again at $D$. Prove that $|AD| = |PB|$.