Since |BD|+|CD|=|BC|=|BE|+|EC|, |BD|^2 + |CD|^2 =|BC|^2 - 2|BD||CD| and |BE|^2 + |EC|^2 =|BC|^2 - 2|BE||EC|. So, 2|DP||DA|+2|BD||CD|= |BC|^2 = 2|EQ||EA| + 2|BE||EC| . Let S,R,T be the second intersection points of AH,AD,AE with circumcircle of BHC, respectively. We know that |DP||DR|=|BD||DC| and |EQ||ET|=|BE||EC| by calculating powers of D and E wrt circle BHC. On the other hand, |DA|+|DR|=|AR| and |EA|+|ET|=|AT|, therefore 2|DP||AR|=|BC|^2=2|EQ||AT|. But, |AP||AR|=|AH||AS|=|AQ||AT|, so |AP|/|DP| = 2|AH||AS|/|BC|^2 = |AQ|/|QE|. From this equality, we can say that PQ // DE which means that PQ // BC. Since B,P,Q,C are concyclic, BPQC is isosceles trapezoid and so |BP|=|CQ|.

It seems too easy for a $4$ and this holds for any circle that passes through $B,C$.Since $BE+CE=BD+CD=BC$,so we have that $BD*CD+DP*DA=BE*CE+EQ*EA$.Now,let $M,N$ be intersections of $AD,AE$ with circumcircle of $BHC$.Now,from PoP we get $EQ*EN=BE*CE$ and $DP*DM=DB*DC$,so summing up we have $EQ*AN=DP*AM$ and use PoP again and we have that $AQ*AN=AP*AM$,so summing again we get $AE*AN=AD*AM$,so $MNDE$ is a cyclic and also $PQMN$ is a cyclic so we get $PQ//DE$ and since $BPQC$ is a cyclic we are finished.

H is useless