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Note that if we could show that $DX\times EY=BD\times DC$, we would be done. Because then, $\dfrac{DX}{BD}=\dfrac{DC}{EY}=\dfrac{BE}{EY}$, and so $\tan \angle XBD=\cot \angle EBY$, implying $\angle XBY$ is a right triangle. Due to symmetey, same could be said for $\angle XCD$, thus making $BXCY$ a cyclic quad.

Here, $M$ is the midpoint of $BC$ and $F$ is the intersection of $AD$ and $OM$. So, $F$ lies on the circumcircle of $\triangle ABC$. WLOG assume $AB<AC$

Now, from an easy application of sine law, we get,

$$\dfrac{AD}{\sin (A+2C)}=\dfrac{DX}{\cos (\dfrac{A}{2}+C)}$$
$$DX=AD\times \dfrac{1}{2\sin (\dfrac{A}{2}+C)}$$

Also, from POP,

$$EY=DY\sin (\dfrac{A}{2}+C)=2DF\sin (\dfrac{A}{2}+C)=\dfrac{2BD\times DC}{AD}\times \sin (\dfrac{A}{2}+C)$$

So, we get, $DX\times EY=BD\times DC$ and we are done.

Our goal is to show that $\angle XCY=90^{\circ}$ (Samely, we can show that $\angle XBY=90^{\circ}$).

To do this, we have to show that $\angle XCD=\angle EYC$ or $\dfrac {EC} {XD}=\dfrac {EY} {CD}$ or $XD \times EY=EC \times CD$.

Let $AD \cap (O)=Q$. As $\triangle AXD \sim \triangle AOQ$, and $OA=OQ$, so, $XA=XD$. Let $XP \perp AD$, then $AP=PD$. Also, $YQ=QD$.

Since $\triangle EYD \sim PDX$,
$\dfrac {DY} {XD}=\dfrac {EY} {DP}$ $\Rightarrow$ $XD \times EY=DY \times DP=\dfrac {DY} {2} \times 2DP=QD \times DA=BD \times DC=EC \times CD$ as desired.

Let , $AD$ meet $( ABC )$ again at $L$ .Then $OL \perp BC$ . $M$ be the midpoint of $BC$ , $K$ be the reflection of $L$ W.R.T $ BC$ and $Y'$ be the reflection of $Y$ W.R.T $ OL$ .$ \angle DKL= \angle KLD=\angle XDA=\angle XAD $ gives $\triangle DAX \sim \triangle KLD \Rightarrow \dfrac {KL}{LD}=\dfrac {DA}{XA} \Rightarrow LD.DA=2ML.XA= EY.XD $
$\Rightarrow BD.DC= LD.DA= EY.XD = DY'.XD$ . So , $ BXCY'$ is cyclic $\Rightarrow BXCYY'$ is cyclic.

Let , $AD$ meet $( ABC )$ again at $L$ .$LM \perp ED \Rightarrow LD=YL=EL.$ Then $\angle EYL=\angle XDA=\angle XAD \Rightarrow \triangle EYL\sim \triangle ADX$. So , $ \dfrac {EY}{AD}=\dfrac {YL}{DX} \Rightarrow EY.DX = AD.YL=AD.LD=BD.DC=EC.DC \Rightarrow \dfrac {EY}{EC}=\dfrac {DC}{DX} .So, \triangle YEC \sim \triangle CDX $. So, $\angle XCY=90^\circ $,similarly$ \angle XBY=90^\circ \Rightarrow BXCY$ is cyclic.

(actually both ideas are same ,written in two different styles )