8. Let $A B C D$ be a convex quadrilateral. Extend line $C D$ past $D$ to meet line $A B$ at $P$ and extend line $C B$ past $B$ to meet line $A D$ at $Q$. Suppose that line $A C$ bisects $\angle B A D$. If $A D=\frac{7}{4}, A P=\frac{21}{2}$, and $A B=\frac{14}{11}$, compute $A Q$.
easily follows.
Denote $\angle B A C=\angle C A D=\gamma, \angle B C A=\alpha, \angle A C D=\beta$. Then we have that by the law of sines, $\frac{A C}{A B}+\frac{A C}{A P}=\frac{\sin (\gamma+\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma-\beta)}{\sin (\beta)}=\frac{\sin (\gamma-\alpha)}{\sin (\alpha)}+\frac{\sin (\gamma+\beta)}{\sin (\beta)}=\frac{A C}{A D}+\frac{A C}{A Q}$ where we have simply used the sine addition formula for the middle step.
Dividing the whole equation by $A C$ gives the desired formula, from which we compute $A Q=\left(\frac{11}{14}+\right.$ $\left.\frac{2}{21}-\frac{4}{7}\right)^{-1}=\frac{42}{13}$