If $L$ is the reflection of $H$ on $D,$ then $DM$ is H-midline of $\triangle BHL$ $\Longrightarrow$ $DX \parallel LB$ and similarly $DY \parallel LC$ $\Longrightarrow$ $\triangle DXY$ and $\triangle LBC$ are homothetic with center $A$ $\Longrightarrow$ $XY \parallel BC.$ Since $\triangle MBD$ is obviously M-isosceles, then it follows that $DBXP$ is an isosceles trapezoid $\Longrightarrow$ $\angle PDB=\angle ABC$ and similarly $\angle QDC=\angle ACB.$ Thus, $\angle PDQ=180^{\circ}-\angle ABC-\angle ACB=\angle BAC=180^{\circ}-\angle PHQ$ $\Longrightarrow$ $PHQD$ is cyclic.

To prove $XY \parallel BC$, let $XY \cap AH = L$. Consider $X(D,H; L, A) \cap BH = Y(D, H; L, A) \cap CH \implies XY \parallel BC$. Now, $MB=MD$ so $PXBD$ is an isosceles trapezoid, hence $DP \cap AB=M'$, the midpoint of $AB$. So, $\angle MDY = 90 = \angle NDX \implies 180 - \angle PDQ = \angle MDN = \angle PHQ$.

Applying Menelaus' theorem twice: \[\frac{AX}{XB} \cdot \frac{BM}{MH} \cdot \frac{HD}{DA} = 1 = \frac{AY}{YC} \cdot \frac{CN}{NH} \cdot \frac{HD}{DA}\] From $M$ and $N$ being the midpoints of $BH$ and $CH$, it follows that $\frac{AX}{XB} = \frac{AY}{YC}$, hence $XY \parallel BC$. The solution then easily follows.

$XY \parallel BC$ because $AXDY$ is cyclic and $PQDH$ is cyclic because both $BDPX$ and $CDQY$ are isosceles trapezoids.

It is a very easy problem. Note that $P$ being the midpoint of hypotenuse $BH$ is the circumcenter of $\odot{BDH}$,so $\angle{BMD}=2\angle{C}$.Also note that $\angle{ABH}=90-\angle{A}$.So $\angle{BXD}=2\angle{C}+\angle{A}-90=90+\angle{C}-\angle{B}$ Similar angle chasing will give $\angle{DYA}=90+\angle{C}-\angle{B}$.Thus $DXAY$ is cyclic,so $\angle{QYD}=\angle{XYD}=\angle{XAD}=\angle{BAD}=90-B=\angle{HCD}=\angle{QCD} \implies QYCD$ is cyclic. So $\angle{PQD}=\angle{YCD}=\angle{C}$.Also note that $\angle{PHD}=\angle{BHD}=\angle{C}$.Thus $PHQD$ is a cyclic quadrilateral as desired.

@sayantanchakraborty: I think you meant M is the midpoint of BH. Also, DXAY cyclic can be shown easily by: Angle XDY= 180 - XDB - YDC = 180 - (90-C) - (90-B) = 180 - A = 180 - XAY.

Lemma : $\angle ADY=\angle B$ $\angle DHB = \angle B$ Now since $N$ is the midpoint of hypotenuse of right triangle $HDC$, $\implies NH=ND$ $\implies $ Our lemma Similarly $\angle XDA= \angle C$ $\implies AXDY$ is concyclic $\implies \angle XYA = \angle C$ $\implies XY||BC$ $PQ||BC$ Now it is easy.

MillenniumFalcon wrote: @sayantanchakraborty: I think you meant M is the midpoint of BH. Quite obviously as it is given in the question

$\triangle XMB$ and $\triangle YNC$ are perspective by line $AD$ so $XY$,$MN$,$BC$ concur $\Rightarrow XY\parallel BC$. $BM=MD$ and $DN=NC$ so $XPDB$ and $YQDC$ are isosceles trapezoids. $\angle DPQ=\angle B$ and $\angle PQD=\angle C\Rightarrow \angle PDQ=\angle A=\pi-\angle BHC$ hence the result.

Solution. Note that $MN \parallel XY$ since the former is a midline of the latter. Moreover, observe that $\bigtriangleup XMB$ and $\bigtriangleup YNC$ are in perspective with respect to $AD$; hence, by Desargues' theorem, $XY\parallel BC$ as well. Since $\bigtriangleup BMD$ is isosceles at $M$, we readily obtain that $XPDB$ is an isosceles trapezoid (hence cyclic); therefore $$\angle DHQ = \angle DHC = \angle ABC = \angle XBD = \angle QPD$$as desired. $\blacksquare$

A really nice problem