Team selection tests for IMO 2018

Denote $R,F$ as the intersections of $AC$ with the angle bisector of $\angle BEC$ and $BE$, respectively. Let $Q$ be a point on $BD$ such that $DQ=DC$, hence $QB=DB-DQ=DB-DC=AD$.



Thus two triangles $CQB$ and $PDA$ are congruent (s.a.s). Note that $AD\perp BE$ then $\angle CAD+\angle AFE=90^{\circ }$. From these, we have

$$\begin{array}{rl}\angle APD& =\angle BCQ=\angle CQD-\angle CBD\\ & =90^{\circ }-\frac{1}{2}\angle BDC-\angle CAD=\left(90^{\circ }-\angle CAD\right)-\frac{1}{2}\angle BEC\\ & =\angle AFE-\angle BER=\angle ARE.\end{array}$$

Hence, we conclude that $DP\parallel ER$. $\square$