8. In the triangle $A B C$, the bisectors of $\angle A$ and $\angle B$ meet at the incentre $I$, the extension of $A I$ meets the circumcircle of triangle $A B C$ at $D$. Let $P$ be the foot of the perpendicular from $B$ onto $A D$, and $Q$ a point on the extension of $A D$ such that $I D=D Q$. Determine the value of $(B Q \times I B) /(B P \times I D)$.
8. Answer: 2 .
Let's prove that $\triangle B P Q$ is similar to $\triangle I B Q$.

First $\angle I A B=\angle C A D=\angle C B D$. As $\angle I B A=\angle I B C$, we have $\angle I A B+\angle I B A=$ $\angle C B D+\angle I B C$ so that $\angle D I B=\angle D B I$, thus $D I=D B=D Q$. This means $\triangle I B Q$ is a right-angled triangle with $\angle I B Q=90^{\circ}$. As $\angle Q$ is a common angle, we thus have $\triangle B P Q$ is similar to $\triangle I B Q$. Hence, $B Q / B P=I Q / I B=2 I D / I B$. Consequently, $(B Q \times I B) /(B P \times I D)=2$.