$EF \cap BC=Z$ $BF \cap EC=W$ $WZ \cap AM=G$ $L_1$ is the radical axis between $(EFBC)$ and $(A)$ (point $A$) By Brocard's theorem $M$ is orthocenter of $\triangle AZW \Rightarrow WZ \perp AM \Rightarrow GH \parallel L_1$ $WZ \cap AM=G \Rightarrow (M,P,G,A)=-1$ $L_1$ meet to $AC$ and $AM$ in $I$ and $K$ respectively. $AI^2=IC.IF$ and $(A,H,F,C)=-1 \Rightarrow I$ is midpoint of $AH \Rightarrow K$ is midpoint of $AG$ $\Rightarrow KP.KM=KA^2=Pot_{(EFBC)}K \Rightarrow K$ and $P$ are inverses points i.e. $L_1$ is the polar of $P$ $O$ is el pole of $XY \Rightarrow O \in L_1 \Rightarrow OX=OY=OA \Rightarrow \angle XOY=2\angle XAY=2\angle XYM$

My solution: Let $ H=BF \cap CE $ and $ T $ be the projection of $ H $ on $ AM $ . Easy to see $ H $ is the orthocenter oF $ \triangle ABC $ . Since $ H $ lie on the polar of $ A $ WRT $ \omega $ , so $ TH $ is the polar of $ A $ WRT $ \omega \Longrightarrow MT \cdot MA=MX^2=MY^2 $ . ... $ (1) $ Since $ A, E, F, H, T $ are concyclic , so $ AP \cdot PT=EP \cdot PF=XP \cdot PY \Longrightarrow A, T, X, Y $ are concyclic . ... $ (2) $ From $ (1) $ and $ (2) $ $ \Longrightarrow MX, MY $ are the tangent of $ \odot (ATXY) \Longrightarrow \angle XAY=\angle XYM $ . Q.E.D

Let $S$ be the other intersection of $AM$ with $(AEF)$. We have by power of a point, $PX\cdot PY=PE\cdot PF=PA\cdot PS$ and so $A,X,S,Y$ lie on a circle. Now, $MX=MY=ME$ and $ME^2=MS\cdot MA$. Letting $O$ be the circumcenter of triangle $AXY$ and $r$ be its radius, we get $MO^2-r^2=MS\cdot MA=MX^2=MY^2$ yielding that $MX,MY$ are tangents to the circle $(AXY)$. The conclusion holds.

Dear Mathlinkers, 1. we can observe that (AEF) is orthogonal to (M) 2. A, T, X, Y are concyclic where T is the second point of intersection of AM with (AEF) and we are done... Sincerely Jean-Louis

We extend $AX$ intersect $\odot M$ at $K$, $XM$ intersect $\odot M$ at $S$, draw $XL$ parallel to $AY$ intersect $\odot M$ at $L$. We have $\angle MXY=\angle MYX$, $\angle KXL=\angle XAY$ so if we want to prove $\angle XAY=\angle XYM$,we just need to prove $\angle KXL=\angle MXY$ namely $\angle KXS=\angle LXY$. We notice $SY \perp XY$,so if $KY \perp XL$ namely $KY \perp AY$ ,we have $\angle TYS=\angle TXY$ which is exactly we want. So we turn to prove $KY \perp AY$ ,for convenience,we donate $AY$ intersect $\odot M$ at $R(R\neq Y)$,so we just need to prove $K,M,R$ is collinear. Then we can put the problem into a brief way: For a fixed circle $\odot M$ and a fixed point $A$,we draw any diameter $BC$ of $\odot M$,$AB,AC$ intersect $\odot M$ at $E,F$,respectively,$AM$ intersect $EF$ at $P$.Prove point $P$ is a fixed point. You can find this problem is equivalent to the orignal problem.(Not hard,for you to think) To solve this problem ,we need to use Menelaus in quadrilateral: For $BEFC$ : We have $\frac{EA}{AB}\frac{BM}{MC}\frac{CA}{AF}\frac{FP}{PE}=1$,it can get $\frac{EP}{PF}=\frac{AC^2}{AB^2}$. For $PFCM$: We have $\frac{PA}{AM}\frac{MB}{BC}\frac{CA}{AF}\frac{FE}{EP}=1$ , we can get $\frac{PA}{AM}=2\frac{AF}{AC}\frac{EP}{EP+PF}=2\frac{AF}{AC}\frac{AC^2}{AC^2+AB^2}=2\frac{AF AC}{AB^2+AC^2}$ Because $AFAC=AM^2-R^2$ is a constant value,and $AM^2=\frac{2AB^2+2AC^2-4R^2}{4}$ namely $AB^2+AC^2=2AM^2+2R^2$ is also a constant value. So we can get: $\frac{PA}{AM}=\frac{AM^2-R^2}{AM^2+R^2}$ is a constant value , which indicates point $P$ is a fixed point on $AM$. Through the process above , we can complete this problem.The rest detailed procedure is for you.

We need to prove $MY$ is tangent to $(AXY)$.Let $AM\cap (AXY)=Z$.Notice that $AP\cdot PZ=XP\cdot PY=EP\cdot PF \implies AEFZ$ cyclic. From 3 Tangents lemma, its easy to see that $MF$ is tangent to $(AEZF)$, $\to MF^2=MZ \cdot MA$ and $MF=MY$ $\implies MY^2=MZ \cdot MA$.

Let $\odot(AXY)\cap AM=T$ Note that $XP\cdot PY=AP\cdot PT=EP\cdot PF$ Which means that $A,E,T,F$is cyclic So $\angle BTM = \angle BEM=\angle EBM$ Hence $\triangle TBM\sim \triangle BAM$ So $BT\cdot TA=BM^{2}=MY^{2}$ Which means that $\angle XAY=\angle XYM$

$\color{red} \textbf{Geo Marabot Solve 3}$ Let $\odot(AXY) \cap AM \equiv T$ $XP\cdot PY=AP\cdot PT=EP\cdot PF \implies AETF$ cyclic. By Three Tangent Lemma, $ME^2 =MT \cdot MA = MY^2 \blacksquare$

Let $AM\cap (AEHF)=T$. Since $XP\cdot PY=EP\cdot PF=AP\cdot PT$, the points $A, X, Y$ and $T$ are concyclic. It's well-known that $ME$ and $MF$ are tangent to $(AEHF)$ at $E$ and $F$, respectively. On the other hand, point $M$ is on radical exis of circles $AEHF$ and $AXYT$. Thus, $MY=ME$ and $MY$ is tangent to $(AXYT)$ implying $\angle XAY=\angle XYM$ as desired.