9. [7] Let $A B C$ be a triangle with $A B=16$ and $A C=5$. Suppose the bisectors of angles $\angle A B C$ and $\angle B C A$ meet at point $P$ in the triangle's interior. Given that $A P=4$, compute $B C$.
Answer:
14
Solution: As the incenter of triangle $A B C$, point $P$ has many properties. Extend $A P$ past $P$ to its intersection with the circumcircle of triangle $A B C$, and call this intersection $M$. Now observe that
$$
\angle P B M=\angle P B C+\angle C B M=\angle P B C+\angle C A M=\beta+\alpha=90-\gamma,
$$
where $\alpha, \beta$, and $\gamma$ are the half-angles of triangle $A B C$. Since
$$
\angle B M P=\angle B M A=\angle B C A=2 \gamma,
$$
it follows that $B M=M P=C M$. Let $Q$ denote the intersection of $A M$ and $B C$, and observe that $\triangle A Q B \sim \triangle C Q M$ and $\triangle A Q C \sim \triangle B Q M$; some easy algebra gives
$$
A M / B C=(A B \cdot A C+B M \cdot C M) /(A C \cdot C M+A B \cdot B M) .
$$

Writing $(a, b, c, d, x)=(B C, A C, A B, M P, A P)$, this is $(x+d) / a=\left(b c+d^{2}\right) /((b+c) d)$. Ptolemy's theorem applied to $A B C D$ gives $a(d+x)=d(b+c)$. Multiplying the two gives $(d+x)^{2}=b c+d^{2}$. We easily solve for $d=\left(b c-x^{2}\right) /(2 x)=8$ and $a=d(b+c) /(d+x)=14$.