Answer: $\square$
$\square$
Solution: Construct parallelogram $B P C Q$. We have $C Q=4, \angle A C Q=90^{\circ}$, and $\angle A B Q=120^{\circ}$. Thus, $A Q=\sqrt{A C^{2}+C Q^{2}}=\sqrt{65}$, so if $x=C P=B Q$, then by Law of Cosine, $x^{2}+6 x+6^{2}=65$. Solving this gives the answer $x=\sqrt{38}-3$.
9. Let $A B C$ be a triangle. Let $X$ be the point on side $\overline{A B}$ such that $\angle B X C=60^{\circ}$. Let $P$ be the point on segment $\overline{C X}$ such that $B P \perp A C$. Given that $A B=6, A C=7$, and $B P=4$, compute $C P$.