......... . Solution ?

We will show that $DEYX$ is an isosceles trapezoid. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Let $\omega$ be the circle with the diameter $AO$. $OX\perp AX, OY\perp AY\Rightarrow X,Y\in \omega$. $\angle OMA=\angle ONA=90^\circ \Rightarrow M,N\in \omega$. $\angle BAX=\angle CAY\Rightarrow \angle MAX=\angle NAY\Rightarrow MX=NY\Rightarrow MN//XY\Rightarrow XY//DE$. We know $MXYN$ and $MDEN$ are isosceles trapezoids. Thus, $$\angle DMX=\angle DMN-\angle XMN=\angle ENM-\angle YNM=ENY$$$$MX=NY$$$$MD=NE$$Then, $\triangle DMX\cong \triangle ENY\Rightarrow DX=EY$. We have $XY//DE$ and $DX=EY$. Hence, $DEYX$ is an isosceles trapezoid.

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Obviously $XY\parallel BC$. And then it's a converse of Radical Axis on $(AXY)$, nine-point circle and circle in question.

Firstly extend $AY\cap BC=Y'$ and $AX\cap BC=X'$ We know that $AD$ and $AO$ are isogonal conjugates, also $AY$ and $AX$ isogonal conjugates $\implies$ $\angle YAD=\angle XAO$ We have $AYOX$ cyclic $\implies$ $\angle AYX= 90-\angle OYX=90-\angle OAX=90-\angle YAD=\angle AY'D$ $\implies$ $XY\parallel BC$ We need to prove that $DEXY$ cyclic so we need to prove $\angle XYD=180-\angle XED=\angle XEX'$ but from parallelism we have $\angle XYD=\angle YDY'$ $\implies$ $\angle YDA=\angle OX'A$ because $OEX'X$ also cyclic so, Let $AD\cap OY=H$ it is sufficent to show that $\triangle AHY'\sim \triangle AOX'$ We have $\triangle HYA\sim \triangle OXA$ $\implies$ $\frac{AH}{AY}=\frac{AO}{AX}$ $\implies$ $\frac{AX}{AY}=\frac{AO}{AH}$, but also from parallelism we have $\frac{AY}{AY'}=\frac{AX}{AX'}$ $\implies$ $\frac{AX}{AY}=\frac{AX'}{AY'}$ from these two we have $\frac{AO}{AH}=\frac{AX'}{AY'}$ $\implies$ $\frac{AO}{AX'}=\frac{AH}{AY'}$ and also $\angle YAD=\angle OAX'$ we have $$\triangle AHY'\sim \triangle AOX'$$