Is there any wrong because its very easy Its clear that the midpoint of XY is also the midpoint of AM and its hypotenuse in the triangle AHM

@above please note that $X \in AB$ and $Y \in AC$

Let $F,E,D$ - midpoints of $AB,AC $and $XY$. Then $ FY \perp AB, EX\perp AC$ so $D$ - midpoint of common hypotenuse of $\triangle AFY,AEY$ so $FD=DE$ and $D$ lies on the perpendicular bisector of $EY$. Let $G$ - midpoint of $EY$ and also it is midpoint of $AM$, so $MG=GH$ and so $G$ lies on perpendicular bisector to $HM$. But $GD \perp HM\to$ $D$ lies on perpendicular bisector to $HM$. So $DH=DM$

Sorry . it's my false Now let Z,E and F be the midpoints of XY ,AB and AC respectively so ZF= ZE (1) (because XF and YE are prependicular on AC and AB respectively) Now. WLOG assume that AC>AB . we have HE=MF and in the same way ME=HF (2), so we conclude that MEF =~HFE ,also HEZ = HEF-ZEF = MFE -ZFE =MFZ (2). By (1),(2)& (3) we deduce that ZHE=~ZMF,so ZM=ZH Q.E.D Its a little bit harder .

Here is an easy complex bash approach. WLOG assume $(ABC)$ is the unit circle, then we know that $$h=\frac{a+b+c-\frac{bc}{a}}{2}, \ m=\frac{b+c}{2}.$$So, the midpoint of $HM$, say $G$, is given by $$g=\frac{a^2+2ab+2ac-bc}{4a}.$$For computing $x$ we know $x+ab\overline{x}=a+b$ (unit circle chord) and $\frac{1}{2}(a+c+x-ac\overline{x})=\frac{a+c}{2}$ (because $AX=XC$ is equivalent to coincidence of the midpoint of $AC$ and the foot from $X$ to $AC$). Thus, we find $$x=\frac{c(a+b)}{b+c}.$$Using symmetry, $$y=\frac{b(a+c)}{b+c}.$$Then the midpoint of $XY$, say $F$, is given by $$f=\frac{ac+ab+2bc}{2(b+c)}.$$We only need to show that the foot from $F$ to $BC$ is $G$. However, $$\frac{1}{2}(b+c+\frac{ac+ab+2bc}{2(b+c)}-bc\frac{\frac{1}{ac}+\frac{2}{bc}+\frac{1}{ab}}{2(\frac{1}{b}+\frac{1}{c})})=\frac{2a(b+c)+a^2-bc}{4a}=g.$$And hence we're done!

Here's my video solution on my channel: https://youtu.be/QexOX_wr8-4