9. $[7]$ Let $A B C$ be a triangle, and $I$ its incenter. Let the incircle of $A B C$ touch side $B C$ at $D$, and let lines $B I$ and $C I$ meet the circle with diameter $A I$ at points $P$ and $Q$, respectively. Given $B I=$ $6, C I=5, D I=3$, determine the value of $(D P / D Q)^{2}$.
Answer:
$\frac{75}{64}$
Let the incircle touch sides $A C$ and $A B$ at $E$ and $F$ respectively. Note that $E$ and $F$ both lie on the circle with diameter $A I$ since $\angle A E I=\angle A F I=90^{\circ}$. The key observation is that $D, E, P$ are collinear. To prove this, suppose that $P$ lies outside the triangle (the other case is analogous), then $\angle P E A=\angle P I A=\angle I B A+\angle I A B=\frac{1}{2}(\angle B+\angle A)=90^{\circ}-\frac{1}{2} \angle C=\angle D E C$, which implies that $D, E, P$ are collinear. Similarly $D, F, Q$ are collinear. Then, by Power of a Point, $D E \cdot D P=D F \cdot D Q$. So $D P / D Q=D F / D E$.
Now we compute $D F / D E$. Note that $D F=2 D B \sin \angle D B I=2 \sqrt{6^{2}-3^{2}}\left(\frac{3}{6}\right)=3 \sqrt{3}$, and $D E=$ $2 D C \sin \angle D C I=2 \sqrt{5^{2}-3^{2}}\left(\frac{3}{5}\right)=\frac{24}{5}$. Therefore, $D F / D E=\frac{5 \sqrt{3}}{8}$.