Notice that $90-\angle ABQ =\angle BAD= \angle BED$ Let $F$ be the intersection of $AC$ and $BE$, since $\angle EFP =90$, $\angle FPE = \angle QPC = 90- \angle BED = \angle ABQ$. In particular $\angle ABQ = \angle QPC$ and $ABQP$ is cyclic.

With altitudes $AK,BL$, $PQ||KL$ (midle line in $\triangle DEH$). Since $ABKL$ - cyclic, so is $ABQP$ (by Reim's).

Join EC, CD,BD,AE ABCD is cyclic, angle BDC=180°-A=B+C angle BCD=angle BAD=90°-B angle ADC=angle B angle BDA=angle C angle CBD°90°-C Since BECD is cyclic, angle CBD= angle DEC= 90°-C Since AECB is cyclic, angle ABE=angle ACE= 90°-A angle CPD=angle ACE+angle CED=180°-(A+C)=B Therefore, angle APQ=180°-angle CPD=180°-B angle APQ+angle ABQ= 180°-B+B=180° So, APBQ is cyclic