Hint

Let $N,M$ be midpoints of $ED$ and $BC$,recpectively. So we have $A-X-N$ are collinear. Since $MD$ and $ME$ are tangents to $(AED)$ we get $AM$ is symmedian in $\triangle AED$. Also since $AY$ is symmedian in $\triangle ABC$, we get $$\angle BAY=\angle CAM=\angle DAM=\angle EAN=\angle EAX\implies A-X-Y \ \text{are collinear.}$$

Let $Z$ be the point where $AX$ cuts $DE$ then by power point we have $ZD=ZE$ so $AX$ is the media on a triangle $AED$ $ED$,$BC$ are anti-// so $AX$ is the symmedian in $ABC$