We shall perform a coordinate bash. Call point $M$ $(a, b)$ and $A$ the origin. Then $B$ is $(0, 2b)$ and $C$ is $(2a, 0)$ The equation of the circumcircle of $ABC$ is $(x-a)^2+(y-b)^2=a^2+b^2$ Notice that the midpoint of arc $BC$ not containing $A$ is the point $(a+b, a+b)$ Let the center of $\omega_1$ be $(g, h)$. Then, $(a-g)^2+(b-h)^2=g^2+(2b-h)^2=h^2$. After some boring calculations we get $g=2a\pm\sqrt{2a^2+2b^2}$ Notice that for any triangle $ABC$, there are two circles through BM tangent to AC, so we have two values. The point $X$ is then $(2a\pm\sqrt{2a^2+2b^2}, 0)$. We can do pretty much the same thing for $\omega_2$ to get $Y$ as $(0, 2b\pm\sqrt{2a^2+2b^2})$. Notice that for the problem statement to be true, we have to be consistent with the $\pm$, so we cannot have a $+$ for the $X$ point but a $-$ for the $Y$ point. Using $X$ and $Y$, we can find line $XY$. Since we already know the midpoint of $BC$, we can just plug in that point into our line, and see if the intersection is the correct one. There are two cases (notice the two possible circles), but fortunately, neither of them require too much calculation. $XY$ is $y=-\frac{2b\pm\sqrt{2a^2+2b^2}}{2a\pm\sqrt{2a^2+2b^2}}+2b\pm\sqrt{2a^2+2b^2}$. We can plug in $x=y=a+b$, and after some algebra we obtain a true statement, so our proof is complete.

The problem does not make sense if $AB = AC$, so I assume $AB \neq AC$. [asy][asy] unitsize(100); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair A, B, C, M, X, Y, P, Q, O; A = dir(130); B = dir(180); C = dir(0); M = (B + C) / 2; X = sqrt(2) * dir(A - C) + C; Y = sqrt(2) * dir(A - B) + B; P = dir(270); Q = dir(90); O = dir(A + Q); draw(circumcircle(B, M, X)^^circumcircle(C, M, Y), gray(0.6)); draw(Y--B--Q--C--X, n_red); draw(Y--Q--X, n_green); draw(A--X^^B--C); draw(P--Y, n_orange + dashed); draw(circumcircle(A, X, Y), n_purple); draw(unitcircle, n_blue); dot(A^^B^^C^^M^^X^^Y^^P^^Q^^O); label("$A$", A, dir(160)); label("$B$", B, dir(200)); label("$C$", C, dir(330)); label("$M$", M, dir(280)); label("$X$", X, dir(20)); label("$Y$", Y, dir(110)); label("$P$", P, dir(270)); label("$Q$", Q, dir(40)); label("$O$", O, dir(160)); [/asy][/asy] WLOG $AB < AC$ and $BC = 2$. Let $P$ and $Q$ be the midpoints of arc $BC$ not containing $A$ and containing $A$, respectively. Let $O$ denote the midpoint of minor arc $AQ$. Observe by PoP that $BY^2 = BM \cdot BC = 2$, whence $BY = \sqrt{2}$ and similarly $CX = \sqrt{2}$. Since $QX = QY = \sqrt{2}$ it follows that $BY = BQ = CX = CQ$, so since $\angle ABQ = \angle ACQ$, $\triangle BQY$ and $\triangle CQX$ are congruent and isoceles. It follows that the perpendicular bisector of $\overline{QY}$ is the bisector of $\angle YBQ$, which is $\overline{BO}$. Similarly $\overline{CO}$ is the perpendicular bisector of $\overline{QX}$; thus $O$ is the circumcenter of $\triangle QXY$. Next, the congruency $\triangle BQY \cong \triangle CQX$ implies \[\angle QYA = \angle QYB = \angle QXC = 180^{\circ} - \angle QXA,\]whence $AXQY$ lie on a circle $\omega$ (with circumcenter $O$). The final claim is that $X$ and $Y$ are the incenter and $P$-excenter of $\triangle APQ$, which will solve the problem. We verify this only for $X$; the verification for $Y$ is similar. First, observe that $C \in \overline{AX}$ is the midpoint of arc $PQ$ not containing $A$, so $\overline{AX}$ bisects $\angle QAP$. Now, note that \[\angle QXA = 180^{\circ} - \angle QXC = 180^{\circ} - \left(90^{\circ} - \frac{1}{2} \angle QCX\right) = 90^{\circ} + \frac{1}{2} \angle QPA\]which means the incenter of $\triangle APQ$ must lie on $\omega$. Since it must also lie on the line $AX$, we conclude that $X$ is the incenter of $\triangle APQ$ as desired.

Apply an Inversion $\Psi$ with center $A$ to get the following problem (I will use the label $X'$ for the inverse of a point $X$): Problem (After inversion): Let $AB'C'$ be a triangle such that $\angle A=90^\circ$. Let $M'$ be the reflection of $A$ WRT $B'C'$. $\omega_1$ is a circle which passes through $B',M'$ and touchs $AC$ at $X'$. $\omega_2$ is a circle which passes through $C',M'$ and touchs $AB$ at $Y'$. The internal bisector of $\angle A$ cuts $BC$ at $D$. Prove that $D$ lies on the circumcircle of $\triangle AX'Y'$. Proof: Let $AC'\cap B'M'=U,AB'\cap C'M'=V$. Note that: $\frac{B'D}{DC'}=\frac{AB'}{AC'}=\frac{AB'}{C'M'}=\frac{UB'}{UC'}\Longrightarrow UD$ bisects $\angle AUB'$ Similarly $VD$ is bisector of $\angle B'VM'$ So $\angle DUX'=\angle DVY' (1)$. On the other hand $UX'^2=UA.UC'=UB'.UA=UY'^2\Longrightarrow UX'=VY' (2)$ and $UD=VD$ (From symmetry). Hence from $(1),(2),(3)\Longrightarrow \triangle UX'D=\triangle VY'D\Longrightarrow Y'D=X'D\Longrightarrow D\in\odot(AX'Y')$. [asy][asy] import graph; size(13.1607359852584cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -17.814556465252437, xmax = 46.34617952000596, ymin = -12.002460930489425, ymax = 18.11174516840608; /* image dimensions */ draw((7.053763172744626,7.656603257387319)--(7.657294332331773,7.277034242897845)--(8.036863346821246,7.880565402484994)--(7.4333321872340985,8.260134416974466)--cycle); draw((8.289196869334935,-0.963123767518571)--(7.875832752621526,-0.38221705868585343)--(7.2949260437888075,-0.7955811753992631)--(7.708290160502218,-1.3764878842319805)--cycle); /* draw figures */ draw(circle((9.556637114817127,3.4984840608820154), 5.213610834073994)); draw(circle((8.450919426224237,2.7116726697758176), 4.155063771069245), blue); draw(circle((9.912043310926615,2.999026475346458), 4.899148278923853), blue); draw(circle((8.213936987931591,5.918095084616983), 2.468702511276209), linetype("2 2")); draw((8.5248009113562,3.4690430822506566)--(10.66298899029792,6.228959008041589)); draw((8.5248009113562,3.4690430822506566)--(5.7648849855652635,5.607231161192375)); draw((-4.48089597794011,15.753155731982314)--(7.4333321872340985,8.260134416974466)); draw((7.4333321872340985,8.260134416974466)--(10.66298899029792,6.228959008041589)); draw((10.66298899029792,6.228959008041589)--(14.768127018076386,3.647181463013054)); draw((14.768127018076386,3.647181463013054)--(7.708290160502218,-1.3764878842319805)); draw((7.708290160502218,-1.3764878842319805)--(-3.7593117383754513,-9.536653807045653)); draw((-3.7593117383754513,-9.536653807045653)--(4.345147211557867,3.3497866587509764)); draw((4.345147211557867,3.3497866587509764)--(5.7648849855652635,5.607231161192375)); draw((5.7648849855652635,5.607231161192375)--(7.4333321872340985,8.260134416974466)); draw((-4.48089597794011,15.753155731982314)--(4.345147211557867,3.3497866587509764)); draw((4.345147211557867,3.3497866587509764)--(7.708290160502218,-1.3764878842319805)); draw((-3.7593117383754513,-9.536653807045653)--(8.5248009113562,3.4690430822506566)); draw((8.5248009113562,3.4690430822506566)--(-4.48089597794011,15.753155731982314)); draw((4.345147211557867,3.3497866587509764)--(7.570811173868158,3.4418232663712427)); draw((7.570811173868158,3.4418232663712427)--(8.5248009113562,3.4690430822506566)); draw((8.5248009113562,3.4690430822506566)--(9.556637114817127,3.4984840608820154)); draw((9.556637114817127,3.4984840608820154)--(14.768127018076386,3.647181463013054)); /* dots and labels */ dot((4.345147211557867,3.3497866587509764),linewidth(2.pt)); label("$B'$", (2.9561794377804023,2.3488404135154646), NE * labelscalefactor); dot((14.768127018076386,3.647181463013054),linewidth(2.pt)); label("$C'$", (15.055637245372347,2.4832788335998193), NE * labelscalefactor); dot((7.4333321872340985,8.260134416974466),linewidth(2.pt)); label("$A$", (7.224599275458671,8.768274972543413), NE * labelscalefactor); dot((7.708290160502218,-1.3764878842319805),linewidth(2.pt)); label("$M'$", (7.123770460395406,-2.793429154711112), NE * labelscalefactor); dot((-3.7593117383754513,-9.536653807045653),linewidth(2.pt)); label("$V$", (-4.437933666859119,-8.876767663528174), NE * labelscalefactor); dot((10.66298899029792,6.228959008041589),linewidth(2.pt)); label("$X'$", (10.787217407694078,6.348383411025024), NE * labelscalefactor); dot((10.66298899029792,6.228959008041589),linewidth(2.pt)); dot((-4.48089597794011,15.753155731982314),linewidth(2.pt)); label("$U$", (-4.94207774217545,16.195997682204023), NE * labelscalefactor); dot((5.7648849855652635,5.607231161192375),linewidth(2.pt)); label("$Y'$", (5.14080376415117,5.6425817055821605), NE * labelscalefactor); dot((5.7648849855652635,5.607231161192375),linewidth(2.pt)); dot((8.5248009113562,3.4690430822506566),linewidth(2.pt)); label("$D$", (8.400935451196778,2.248011598452198), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

let Q be a midpoint of arc BAC.and XY meet arc BAC at Z and arc BC at T we have CX=BY and BQ=CQ and <ACQ=<ABQ so BYQ=CXQ so <CXQ=<BYQ so YAXQ is cyclic. CX=BY=BQ=CQ=2^1/2 * BM so bisector perpendicular of YQ and QX is bisector of <YBQ and <XCQ so Z is center of circumcircle of YAXQ. so arc AO=arc OQ we want to show arc TC=arc QC we have OA=OX so <OAX=<OXA so arc OA +arc TC =arc OQ+arc QC so arc TC=arc QC

Assume, without loss of generality, that $AC>AB$. Let the bisector of $\angle{A}$ intersect $BC$ and the circumcircle of $\triangle{ABC}$ at $K$ and $L$ respectively. Also, let the parallel from $L$ to $BC$ meet $AB$ and $AC$ at $D$ and $E$ respectively. We have that: $$\frac{AB}{BD}=\frac{AC}{CE}=\frac{AK}{KL}=\frac{AB*AC}{LB*LC}=\frac{2AB*AC}{BC^2}$$So, we have that $BD=\frac{BC^2}{2AC}$ and $CE=\frac{BC^2}{2AB}$. The tangency condition gives that $CX^2=CM*CB=\frac{BC^2}{2} \Longleftrightarrow CX=\frac{BC}{\sqrt{2}}$. Similarly, we have $BY=\frac{BC}{\sqrt{2}}$. Now, we compute: $$YA=BY-BA=\frac{BC}{\sqrt{2}}-AB$$$$YD=YB+BD=\frac{BC}{\sqrt{2}}+\frac{BC^2}{2AC}=\frac{BC(BC+\sqrt{2}AC)}{2AC}$$$$XA=CA-\frac{BC}{\sqrt{2}}$$$$XE=XC+CE=\frac{BC(BC+\sqrt{2}AB)}{2AB}$$$$\frac{LD}{LE}=\frac{KB}{KC}=\frac{AB}{AC}$$From Menelau's Theorem applied in $\triangle{ADE}$, it is sufficient to prove that: $$\frac{YA}{YD}*\frac{LD}{LE}*\frac{XE}{XA}=1 \Longleftrightarrow$$$$\frac{\frac{BC}{\sqrt{2}}-AB}{\frac{BC(BC+\sqrt{2}AC)}{2AC}}*\frac{AB}{AC}*\frac{\frac{BC(BC+\sqrt{2}AB)}{2AB}}{CA-\frac{BC}{\sqrt{2}}}=1\Longleftrightarrow$$$$\frac{\frac{BC}{\sqrt{2}}-AB}{BC+\sqrt{2}AC}*\frac{BC+\sqrt{2}AB}{CA-\frac{BC}{\sqrt{2}}}=1 \Longleftrightarrow$$$$\frac{(BC-\sqrt{2}AB)*(BC+\sqrt{2}AB)}{\sqrt{2}}=\frac{(\sqrt{2}CA-BC)*(\sqrt{2}CA+BC)}{\sqrt{2}} \Longleftrightarrow$$$$BC^2-2AB^2=2AC^2-BC^2 \Longleftrightarrow 2BC^2=2(AB^2+AC^2)$$which is true by Pythagoras since $\angle{A}=90^\circ$.

Let $S$ be midpoint of arc $BC$. First Note that $BY^2 = BM.BC = CM.CB = CX^2 \implies CX = BY = \frac{BC}{\sqrt{2}} = BS = SC$. Let $SX$ meet $AB$ at $Y'$. $\angle XY'A = \angle 90 - \angle AXY' = \angle 90 - \angle SXC = \angle 90 - \angle XSC = \angle XSB \implies SB = BY' \implies BY = BY'$ so $Y'$ is $Y$ so $Y,X,S$ are collinear.