Pre-IMO 2017 Mock Exam

Note that $\angle PDQ=180^{\circ }-\angle BDP-\angle CDQ=180^{\circ }-\frac{1}{2}\angle BDE-\frac{1}{2}\angle CDF=180^{\circ }-\frac{A}{2}-\frac{A}{2}=180^{\circ }-A$. This implies $A,P,D,Q$ are concyclic. Hence, $\angle APQ=\angle ADQ=180^{\circ }-\angle DAQ-\angle ACD-\angle CDQ=180^{\circ }-\frac{B-C}{2}-C-\frac{A}{2}=90^{\circ }$. $\square$

---

Alternative solution.

From the concyclic points, we have $\angle DFC=\angle DBE$ and $\angle DCF=\angle DEB$. These show $△DFC$ and $△DBE$ are directly similar. Thus, there exists a spiral similarity with centre $D$ mapping $△DFC$ to $△DBE$. Note that the image of $Q$ is $P$ under this transformation since they are the corresponding points on sides $CF$ and $EB$ respectively. It follows that $△DPQ$ and $△DBF$ are directly similar. Then $\angle (PQ,AB)=\angle (PQ,BF)+\angle FBA=\angle (PD,BD)+(B-\angle CBF)=\frac{1}{2}\angle EDB+B-\angle DAC=\frac{A}{2}+B-\frac{B-C}{2}=90^{\circ }$. $\square$