Question 8
In triangle $ABC$, the bisectors of $\angle A$ and $\angle B$ meet at the incentre $I$. The extension of $AI$ meets the circumcircle of $ABC$ at $D$. Let $P$ be the foot of the perpendicular from $B$ onto $AD$, and $Q$ a point on the extension of $AD$ such that $ID=DQ$. Determine $\dfrac{BQ\cdot IB}{BP\cdot ID}$.