China Western Mathematical Olympiad

On the ray $HD$, mark a point $M$ such that $HD=DM$. Join the segments $BM,CM,BH$ and $CH$. As $D$ is the midpoint of $BC$, the quadrilateral $BHCM$ is a parallelogram, and so $$\angle BMC=\angle BHC=180^{\circ }-\angle BAC.$$ Hence, $$\angle BMC+\angle BAC=180^{\circ },$$ and the point $M$ lies on the circumcircle of $△ABC$. Join the segments $PB,PC,PE$ and $PF$. It follows from $AE=AF$ that



$$\angle BFH=\angle CEH.\stackrel{◯}{1}$$ As $H$ is the orthocenter of $△ABC$, $$\angle HBF=90^{\circ }-\angle BAC=\angle HCE.\stackrel{◯}{2}$$ From ① and ②, one has $△BFH\sim △CEH$, so $$\frac{BF}{BH}=\frac{CE}{CM}.$$ As $BHCM$ is a parallelogram, $BH=CM,CH=BM$, and hence $$\frac{BF}{CM}=\frac{CE}{BM}.\stackrel{◯}{3}$$ And $D$ is the midpoint of $BC$. Thus, $S_{△PBM}=S_{△PCM}$, and so $$\frac{1}{2}BP\times BM\times \sin \angle MBP=\frac{1}{2}CP\times CM\times \sin \angle MCP.$$ It follows from $\angle MBP+\angle MCP=180^{\circ }$ that $$BP\times BM=CP\times CM.\stackrel{◯}{4}$$ With ③ and ④, one has $\frac{BF}{BP}=\frac{CE}{CP}$. From $\angle PBF=\angle PCE$, one has $△PBF\sim △PCE$, and hence $\angle PFB=\angle PEC$, and therefore $\angle PFA=\angle PEA$. Consequently $P,A,E$ and $F$ are concyclic.