APMO

Let $\ell$ be the radical axis of circles $BQX$ and $CPX$. Since $X$ and $Y$ are on $\ell$, it is sufficient to show that $A$ is on $\ell$. Let line $AX$ intersect segments $BC$ and $DE$ at $Z$ and $Z^{\prime }$, respectively. Then it is sufficient to show that $Z$ is on $\ell$. By $BC\parallel DE$, we obtain $$\frac{BZ}{ZC}=\frac{D{Z}^{\prime }}{Z^{\prime }E}=\frac{PZ}{ZQ},$$ thus $BZ\cdot QZ=CZ\cdot PZ$, which implies that $Z$ is on $\ell$.

---

Alternative solution.

Let circle $BQX$ intersect line $AB$ at a point $S$ which is different from $B$. Then $\angle DEX=\angle XQC=\angle BSX$, thus $S$ is on circle $DEX$. Similarly, let circle $CPX$ intersect line $AC$ at a point $T$ which is different from $C$. Then $T$ is on circle $DEX$. The power of $A$ with respect to the circle $DEX$ is $AS\cdot AD=AT\cdot AE$. Since $\frac{AD}{AB}=\frac{AE}{AC}$, $AS\cdot AB=AT\cdot AC$. Then $A$ is in the radical axis of circles $BQX$ and $CPX$, which implies that three points $A$, $X$ and $Y$ are collinear.

---

Alternative solution.

Consider the (direct) homothety that takes triangle $ADE$ to triangle $ABC$, and let $Y^{\prime }$ be the image of $Y$ under this homothety; in other words, let $Y^{\prime }$ be the intersection of the line parallel to $BY$ through $D$ and the line parallel to $CY$ through $E$. The homothety implies that $A$, $Y$, and $Y^{\prime }$ are collinear, and that $\angle D{Y}^{\prime }E=\angle BYC$. Since $BQXY$ and $CPXY$ are cyclic, $$\angle D{Y}^{\prime }E=\angle BYC=\angle BYX+\angle XYC=\angle XQP+\angle XPQ=180^{\circ }-\angle PXQ=180^{\circ }-\angle DXE,$$ which implies that $D{Y}^{\prime }EX$ is cyclic. Therefore $$\angle D{Y}^{\prime }X=\angle DEX=\angle PQX=\angle BYX,$$ which, combined with $D{Y}^{\prime }\parallel BY$, implies $Y^{\prime }X\parallel YX$. This proves that $X$, $Y$, and $Y^{\prime }$ are collinear, which in turn shows that $A$, $X$, and $Y$ are collinear.