It's well-known that the circle $\omega$ with center $M$ and radius $MB=MC$ goes through the incenter $I$ and A- excenter $I_a$ of $\triangle ABC.$ Since $(A,D,I,I_a)=-1,$ it follows that external bisector $\ell_a$ of $\angle BAC$ is the polar of $D$ with respect to $\omega.$ If $XY$ cuts $\ell_a$ at $Z,$ then cross ratio $(Z,D,X,Y)$ is harmonic. Together with $DA \perp AZ,$ we deduce that $AD,\ell_a$ are the internal and external bisector of $\angle XAY.$

Let $AE$ and $AF$ be tangents to $\omega$ ($E$ and $F$ are the tangency points). It is easy to see that $D,E,F$ are collinear. From here, triangles $AEF$, $ABC$, and $AXY$ share the same incenter, hence $AD$ bisects $\angle XAY$.

Let $ G_1 $ be the circumcircle of $ \triangle ABC $. Let $ G_2 $ be the circumcircle of $ \triangle XYM $. Since $ \angle XYM =\angle YXM $ as $ MX = MY $. Thus it suffices to prove that the 4 points $ A,Y,M,X $ are concyclic. (as then $ \angle XAD = \angle MAY $) If $ A,Y,M,X $ are not concyclic, then suppose $ G_1 $ and $ G_2 $ intersect at $ Q $. ($ Q $ not equal to $ A $ ) Now, by radical axis theorem, $ MQ, BC, XY $ are concurrent which cannot be true as $ MA, BC, XY $ are concurrent. Hence $ Q = A $ and $ A,Y,M,X $ are concyclic and we're done.

Dear Mathlinkers, 1. According to Monge's theorem, A, M, X and Y are concyclic 2. MX=MY and we are done... Sincerely Jean-Louis

Sorry for reviving, but today I found another solution to this. Let $R$ be the radius of ($M$). Construct the tangent lines of ($M$) passing through $A$, called $AE,AF$ ($E,F$ lie on ($O$)). Triangle $MBD,MAB$ are congruent ($\widehat{BMD}=\widehat{AMB}, \widehat{MBD}=\widehat{MAC}=\widehat{MAB}$) $\Rightarrow\frac{MD}{MB}=\frac{MB}{MA}$ $\Rightarrow MD.MA=MB^2=R^2$ $\Rightarrow D$ is the image of $A$ under the inversion ($M,R^2$). $\Rightarrow D$ lies on $EF$ and $D$ is the midpoint of $EF$. So now it remains to prove that $AX,AY$ cut $EF$ at two points that is symmetry to each other wrt $D$. ($1$) Let $AY$ cuts ($M$) at a second point $G$. $GD$ cuts ($M$) at a second point $H$. Let $AY,AX$ cuts $EF$ at $U,V$ respectively.By Butterfly's theorem, we conclude that $U,V$ are symmetric wrt ($D$) We have $DU$ is the angle bisector of $\widehat{GDI}$ (well-known) and since $DA$ is perpendicular to $DU$, we conclude that ($A,U,G,Y$)=$-1$. $\Rightarrow D(A,U,G,Y)=-1$ $\Rightarrow D(A,V,X,H)=-1$ $\Rightarrow A,V,X,H$ are collinear. So ($1$) has been proved.The proof is completed. Q,E,D

$D$ lies on the radical axis of the $2$ circles, $DA\times DM=DB\times DC=DX\times DY$ so $A,Y,M,X$ concyclic, since $MX=MY$ we are done.

<BAM = <CAM or, <BCM = <CBM; or, BM = CM so, w passes through C. A,B,C,M cyclic by power of point, AD*MD = BD*CD. B,C,X,Y cyclic so, BD*CD = XD*YD; now, XD*YD = AD*MD A,M,X,Y cyclic so, <XAM = <XYM = <YXM = <YAM

By power of point in $\omega$ we have $\overline{DB} \times \overline{DC} = \overline{DY} \times \overline{DX}$. And by applying power point again in $(ABC)$ we have $\overline{DB} \times \overline{DC} = \overline{DM} \times \overline{DA}$. So we gain that $\overline{DY} \times \overline{DX}=\overline{DM} \times \overline{DA}$. So, $A,M,X,Y$ are concyclic. And it is given that $M$ is the circumcenter of $\omega$. So $\overline{MX}=\overline{MY}$. So $MA$ bisects $\angle XAY$. $\mathbb Q. \exists. \mathbb D.$

Attachments:

It's easy to find out that what we need is to prove AXMY is cyclic. ABMC is cyclic ---> AD.DM = BD.CD (1) BXCY is cyclic ---> BD.DC = XD.YD (2) By 1 and 2 we have AD.DM = XD.YD so AXMY is cyclic. and we're Done

The Shooting Lemma yields $$MA \cdot MD = MB^2 = MX^2 = MY^2.$$Because $D, X, Y$ are collinear, inverting about $\omega$ implies $AXMY$ is cyclic. But $MX = MY$, so the result follows easily. $\blacksquare$

That's enough to prove that $AXMY$ is cyclic. $AXMY$ is cyclic $\iff$ $XD\cdot DY=DM\cdot DA$ $\iff$ $BD\cdot DC=AD\cdot DM$ which is true.