Warning

Let's start with defining points. Let $Z$ be $A-\text{excenter}$ of $\triangle ABC$, $I$ is incenter of $\triangle ABC$, $L=(ABC)\cap AI$,$R=MX\cap AB$, $P=MX\cap AC$,$Q=MY\cap AB$,$S=MY\cap AC$. Now let $P'$ be a point on $AC$ such that $IP'||BC$. Some calculations gives that $\frac{MB}{IP'}=\frac{BX}{IX} \implies M-P'-X$ are collinear $\implies P'=P$. So $IP||BC$. Similarly, $IQ||BC$. So we get $PQ||BC$ annd $I\cap PQ$. Some easy angle chase gives that $IPCL$ is kite. So $LP\perp IC\perp ZC \implies LP||ZC$. Similarly $LQ||ZB$. So $L$ is $A-\text{excenter}$ of $\triangle APQ$ $(\star)$. In a same way let $S'$ be a point on $AC$ such that $ZS'||BC$. Again some caluculations gives that $\frac{YB}{YZ}=\frac{BM}{ZS'} \implies Y-M-S'$ are collinear $\implies S'=S$. So $ZS||BC$. Similarly $ZR||BC$. So $RS||BC$ and $Z\in RS$. Easy angle chase gives that $BRZL$ is kite. So $RL\perp BZ\perp BI \implies RL||BI$. Similarly, $SL||CI$. So $L$ is incenter of $\triangle ARS (\star \star)$. Finally from $(\star)$ and $(\star \star)$ we get that there is circle with center $L$ inscribed in $PQRS$. So we are done!