HongKong 2022-23 IMO Selection Tests

Note that $\sin \angle DAB=\sin \angle DCB$ as the two angles are supplementary. Using $[XYZ]$ to denote the area of $XYZ$, we have $$\frac{AE}{EC}=\frac{[DAB]}{[DCB]}=\frac{\frac{1}{2}\cdot AD\cdot AB}{\frac{1}{2}\cdot DC\cdot BC}=\frac{AB}{BC}\cdot \frac{AD}{DC}(1)$$ By the angle bisector theorem, we have $$\frac{AB}{BC}=\frac{AP}{PC}\text{and}\frac{AD}{DC}=\frac{AQ}{QC}$$ It follows from (1) that $$\frac{AE}{EC}=\frac{AP}{PC}\cdot \frac{AQ}{QC}$$ Setting $PC=x$ and using the given side lengths, this becomes $$\frac{4}{5+x}=\frac{9}{x}\cdot \frac{6}{3+x}$$ Solving gives $x=15$ or $x=-\frac{9}{2}$, and of course the latter is rejected.