$BE=BA,BH=HA$ since $ (ABC),(BCH),(BAH)$ are congruent and $\angle BHE =\pi - \angle BCA$ hence $BH$ is the bisector of $AE$ idem $CH$ is the bisector of $AD$ besides from the power of $P,Q$ we deduce $AXDE,AYDE$ are cyclic so they all lie on the circle of center $H$ and radius $AH$ ;from $PX.PA=PD.PE$ and $P $is in the $EA$ -bisector we get $XA=DE$ idem from $QY.QA=QD.QE$ and $Q $ is in the $EA$ -bisector we get $YA=DE$ so $HA$ is the bisector of $XY$ therefore $XY\parallel BC$ RH HAS

We can show that $ABC,BCH,BAH,BCH$ has same circumradius. $\angle BEA= \angle A$ so $BH$ is perpendicular bisector of $AE$ so $PH$ is axis of symmetry, and so $AH=HE$ and $AX=DE$ Same way for $AH=HD$ and $DE=AY$ $\angle XAH= \angle DEH= \angle EDH=\angle YAH$ and so $AH$ is perpendicular bisector for $XY$ $\to XY\parallel BC$

Claim : $A,D,E,X,Y$ are concyclic. Proof : Observe that reflection of $E$ in $BH$ is $A$. Now since, $D,B,H,E$ are concyclic, $X$ is the reflection of $D$ in $BH$. $\implies X \in \odot {ADE}$ Similarly, $Y \in \odot {ADE}$ and hence the proof. Also, $B,H,E,C,D$ are concylic. Hence by Reim's theorem, $BC || XY$

Let $H_B,H_C$ be foot of perpendicular from $B,C$. then simple angle chasing shows $\implies$ $BH_B,CH_C$ are perpendicular bisector of $AE$ $,AD$ $\implies$ $DQ$ $=$ $AQ$ and $AP$ $=$ $AE$ $$DQ ~~~ \cdot ~~~ QE=HQ ~~~ \cdot ~~~ QC=AQ ~~~ \cdot ~~~ QY=DQ ~~~ \cdot ~~~ QY \implies QE=QY \implies D-Y-C$$Similarly, $B-X-E$ and isosceles trapeziums show $\implies$ $A,D,X,Y,E$ concyclic, Now Apply Reim's theorem and done!

The Key Observations are:

Points $A,D,E,X,Y$ are concyclic. Proof

Points $A,H,D,P$ are concylic. Proof

Points $A,H,E,Q$ are concyclic. Proof

[asy][asy] size(200); pair A=dir(115),B=dir(-150),C=dir(-30),H=A+B+C,D=2*foot(H,A,B)-A,E=2*foot(H,A,C)-A,P=extension(D,E,H,B),Q=extension(D,E,H,C),X=2*foot(H,A,P)-A,Y=2*foot(H,A,Q)-A; draw(circumcircle(A,D,E)^^arc(circumcenter(A,H,P),P,A)^^arc(circumcenter(A,H,Q),A,E),orange); draw(arc(circumcenter(B,H,C),C,D)^^arc(circumcenter(A,B,H),X,A)^^arc(circumcenter(A,H,C),A,C),purple); dot("$A$",A,dir(A)); dot("$B$",B,dir(150)); dot("$C$",C,dir(C)); dot("$H$",H,dir(H)); dot("$D$",D,dir(-90)); dot("$E$",E); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(70)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); draw(B--C^^X--Y,brown); draw(C--A--D--E--P,fuchsia); draw(P--foot(H,A,C)^^C--foot(H,A,B),green); draw(P--A--Y,cyan); [/asy][/asy] With the above three observations in mind, the finish is just angle chase: \begin{align*} \angle (\overline{XY},\overline{AB}) &= \angle AXY + \angle XAD = \frac{1}{2} \left( \widehat{AE} + \widehat{XD} + \widehat{YE} \right) = \angle ADE + (\angle EDY + \angle BYD) = \angle C + \angle (\overline{DE},\overline{BC}) \\ &= \angle C + (\angle ABC - \angle ADE) = \angle C + (\angle B - \angle C) = \angle B = \angle ABC \end{align*}

$\textcolor{blue}{\text{Claim 1.}\hspace{0.1cm} AXDYE \hspace{0.1cm} \text{cyclic}}$ Proof: By PoP in $AHYC$ and $HBDCE$ we have \[AQ\cdot QY= HQ\cdot QC= DQ\cdot QE\]so $AEYD$ cyclic, similarly $AEDX$ cyclic. $\textcolor{blue}{\text{Claim 2.}\hspace{0.1cm} YE\parallel AB \hspace{0.1cm} \text{and} \hspace{0.1cm} XD\parallel AC}$ Proof: To prove $YE\parallel AB$ it suffices to prove $CA=CD$ as $AEYD$ cyclic, now note \[\angle HCD=90^{\circ}-\angle ADC=90^{\circ}-(180^{\circ}-\angle BHC)=(180^\circ -\angle HBC-\angle HCB)-90^\circ=\angle HCA\]as desired, similarly $XD\parallel AC$ $\textcolor{blue}{\text{Claim 3.}\hspace{0.1cm} \text{The tangent throught}\hspace{0.1cm}A \hspace{0.1cm} \text{in} \hspace{0.1cm} AXDYE\hspace{0.1cm} \text{is parallel to} \hspace{0.1cm} BC}$ Proof: This is true by Reim in $AXDYE$ and $BECD$. Finally, by Pascal in $AADXYE$ we obtain $AA\cap XY, AD\cap YE, DX\cap EA$ must be the line on infinity (by Claim 2) hence by Claim 3, $XY$ is parallel to $BC$, as desired.