let $Z$ intersection of altitudes. So $\angle CXM = 90^\circ$ let $XM\cap \omega=R$ and $CR$ diameter. So $\angle CBR =\angle CAR = 90^\circ$. So $AZ \parallel RB$ and $AR \parallel BZ$. So $AZBR$ parallelogram and $ZR$ passes from $M$ so $ZR\cap \omega=X$. Let $AB\cap XC=T$. Because of $MZ,CZ$ are altitudes in $CTM$, $TZ$ perpendicular to $CY$. Let $TZ\cap CY=S$. Power in $\omega$ we have $TA.TB=TX.TC$ and power in $(CXZ)$ we have $TZ.TS=TX.TC$ so $TZ.TS=TA.TB=TX.TC$. So $ABSZ$ cyclic. Because of $CTHS$ cyclic we have $\angle TCH = \angle HST$. Also $AZSB$ cyclic so $\angle ZSA = \angle ZBA = \angle ACZ$. So $\angle HYA = \angle CXA = \angle TCH - \angle ACZ = \angle TSH - \angle ZSA = \angle ASH $. Because of $HXCM,HTCS$ cyclic we have $\angle YHM = \angle XHT = \angle SCX = \angle SHM $. So because of $\angle HYA = \angle ASH $ and $\angle YHM = \angle SHM$. So we have $S$ and $Y$ symetric respect to $AB$. power lines of $CXZ$,$ABC$,$ABPQ$ intersect one point. Where $BZ\cap AC=Q$. And $AZ\cap BC=P$. So $P,Q,T$ linear. So we have because of $AP,BQ,CH$ intersect at $Z$ $T,A,H,B$ harmonic. So $HM.MT=MA^2$. So $HM.MT=MS.MC=MA^2$. Because of $CSHT$ cyclic $HM.MT=MS.MC$. Because of $MS.MC=MA^2$ we have $\angle SAB = \angle ACM$ and $\angle SBA = \angle MCB$. Because of $Y,S$ symetric $\angle YAB = \angle ACM$ and $\angle YBA = \angle BCM$. Because of $\angle YAB = \angle ACM$,$\angle YBA = \angle BCM$ and $AYBC$ cyclic we have $\angle YCH = \angle ACY - \angle ACH = \angle ABY - \angle DCB = \angle BCM - \angle DCB = \angle DCM$. Because of $CR$ diameter $D$ is on $CR$. So we have $\angle YCD = \angle YCM + \angle DCM = \angle YCM + \angle YCH = \angle HCS$. Because of $S,Y$ symetric and $THSC$ cyclic we have $\angle YTD = \angle STM = \angle SCH = \angle YCD$ so $TDYC$ cyclic. So we had $\angle XCY = \angle HDY$. If we take $l=LY$ tangent to $(HYD)$ we have $\angle LYX = \angle HDY$. Let $L$ closer to $H$. So we have $\angle LYX = \angle XCY$. So we have $l$ tangent to $(ABC)$ but we didn't use isosceles. So we have result.

It's well-known that $CY$ is $C-$symmedian of triangle $CAB,$ combine with $(CHD)$ is tangent to $(O):$ call $T$ the intersection of tangent at $Y$ of $(O)$ and common tangent at $A$ of $(CHD)$ and $(O)$ with $BC$ Then $TY^2=TA^2=TH.TD,$ so $(YHD)$ is tangent to $(O)$