Let $E,F$ be the projections of $D$ ond $BA,BC$, respectively, and let $X$ be on $BA$ such that $FX \parallel AD$, and $Y$ be on $BC$ such that $EY \parallel DC$. We will prove these two points work. The problem splits into two parts. Part 1: $XY$ is perpendicular to $BD$. Firstly, note that $\angle EXF=\angle EAD=180^\circ-\angle BAD=180^\circ-\angle BCD=\angle DCF=\angle EYF,$ and so points $X,Y,E,F$ are concyclic. Hence, in light of $EBFD$ being cyclic, too, $\angle BXY+\angle XBD=\angle EFB+\angle XBD=\angle EFB+\angle EFD=90^\circ,$ and so $XY$ is perpendicular to $BD,$ as desired. Part 2: $MX=MY$ holds true. We have the following 2 Claims. Claim 1: $ME=MF$ holds true. Proof: Note that, due to the Law of Cosines, $ME^2=AM^2+AE^2-2AM \cdot AE\cos \angle CAE=AM^2+AE^2-AE \cdot AC\cos \angle CAE,$ and similarly $MF^2=MC^2+CF^2-AC \cdot CF \cos \angle ACF$ Now, if $P,Q$ are the projections of $C,A$ on $BE,BF$, respectively, then $ME^2=AM^2+AE^2-AE \cdot AC\cos \angle CAE=AM^2+AE^2-AE \cdot AP=AM^2+EA \cdot EP$ and similarly $MF^2=MC^2+CF^2-AC \cdot CF \cos \angle ACF=MC^2+CF^2+CQ \cdot CF=MC^2 +CF \cdot FQ$ Hence, it is enough to prove that $EA \cdot EP=FC \cdot FQ,$ that is $EA/FC=FQ/EP$. Now, note that triangles $EAD,FCD$ are similar, and so $EA/FC=AD/DC$. Lastly, $\angle(AD,FQ)=90^\circ-\angle QAD=\angle ADF-90^\circ=\angle EDC-90^\circ=90^\circ-\angle ACD=\angle(AE,DC),$ hence $FQ/EP=(AD\cos \angle(AD,FQ))/(DC\cos \angle(AE,DC))=AD/DC=EA/FC,$ as desired $\blacksquare$ Claim 2: $\angle MEF=\angle MFE=\angle EDA=\angle FDC$. Proof: Let $\angle MEF=\angle MFE=x$ (those angles are equal due to Claim 1), and $\angle EDA=\angle FDC=y$. Then, $\angle AEM=\angle AEF-\angle MEF=\angle BDF-x=\angle BDC+y-x,$ and similarly $\angle MDC=\angle ADB+y-x$. Moreover, $\sin \angle MDC/\sin \angle AEM=(AM/\sin \angle AEM)/(MC/\sin \angle MDC)=$ $=(EM/\sin \angle EAC)/(MF/\sin \angle ACF)=\sin \angle ACF/\sin \angle EAC=BA/BC,$ and $\sin \angle ADB/\sin \angle BDC=BA/BC,$ as triangles $BAD,BDC$ have $\angle BAD=\angle BCD$. Hence, the two ratios $\sin \angle MDC/\sin \angle AEM$ and $\sin \angle ADB/\sin \angle BDC$ are equal and from the above angle equalities we obtain $\angle MDC-\angle AEM=(\angle ADB+y-x)-(\angle BDC+y-x)=\angle ADB-\angle BDC,$ hence by a well-known lemma $\angle MDC=\angle ADB, \angle AEM=\angle BDC$. Therefore, $\angle ADB+y-x=\angle MDC=\angle ADB,$ and so $x=y,$ as claimed $\blacksquare$ Now, back to the problem, by Claim 2 we obtain $\angle EMF=180^\circ-2\angle MEF=180^\circ-2\angle EDA=2\angle EAD=2\angle EXF,$ hence $M$ is the circumcenter of $(EXYF),$ and so $MX=MY$, as desired.

Yay, that's one more my problem on the international contest. On my opinion, this problem is much better than IMO 2023 P2 so I encourage you all to try it.

Solution sketch (modulo some angle chasing details). Let $D'$ and $B'$ be the reflections of $D$ and $B$ across $M$. Let $X'$ and $Y'$ be the point on $BA$ and $BC$ such that $D'X' = D'Y' = D'B$. Using the parallelogram isogonality lemma (which implies that $BD$ and $BD'$ are isogonal), we can angle chase to get that $B'AX'D'$ and $B'CY'D'$ are concyclic. Angle chase more to get that $B'X' = B'Y'$, $X'Y'\perp BD$, and that $\angle X'B'Y' = \angle ADC - \angle BDC$. Finally, we can let $X$ and $Y$ be midpoints of $BX'$ and $BY'$ and notice that $\triangle MXY$ and $\triangle B'X'Y'$ are homothetic with ratio $1:2$.

Let the reflection of $D$ through $M$ be $E$.. Let $X,Y \in AB,BC$ st $\angle EXB=\angle EYB=90$ .It's well-known that $BE,BD$ are isogonal wrt $\angle ABC$ so since $BE$ is the diameter of $(BXY)$ , we have $XY \perp BD$. Now let $Z \in AC$ st $EZ \perp AC$ . one can easily see that the isogonal conjugate of $E$ in $\triangle ABC$ lies on the perpendicular bisector of $AC$ (since $\angle EAB= \angle ECB$) So $M$ lies on the six-point circle of $E$ in $\triangle ABC$ and $M \in (XYZ)$ so $\angle XMY=\angle XZY=\angle XZE+\angle YZE=\angle XAE+\angle YCE=\angle AEC-\angle ABC=\angle ADC-\angle ABC$ and finally , let $X',Y' \in AB,BC$ st $\angle DX'B=\angle DY'B=90$. since $EDX'X$ was a right trapezoid , $MX=MX'$ and $MY=MY'$ and by easy angle chasing , $XX'Y'Y$ is cyclic. So $M$ is the center and $MX=MY$ and we're done.

Let be $E$ the symmetric of $B$ with relation to $M$. Notice that $ABCE$ is a parallelogram and $\angle EAB=\angle ECB$. As $\angle BAD=\angle BCD$, we get $\angle EAD=\angle ECD=\alpha$. Also, $\angle AEC=\angle ABC$. The condition $\angle ABC<\angle ADC$ implies that point $D$ is inside the triangle $EAC$. We have $\angle ADC=\angle AEC+2\alpha$, i.e., $$2\alpha=\angle ADC-\angle ABC=\angle XMY.$$Let be $X'$ and $Y'$ the reflections of $X$ and $Y$, respectively, across $M$. Notice that there exist points $X$ and $Y$ satisfying the problem if, and only if, points $X'$ and $Y'$ on the segments $EC$ and $EA$ satisfy $MX'=MY'$, $X'Y'\bot BD$ and $\angle X'MY'=2\alpha $. Let be $N$ the midpoint of segment $DE$. So, $MN$ // $BD$. Then, $X'Y'\bot BD$ iff $X'Y'\bot MN$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.188125298038734, xmax = 5.087058573682681, ymin = -14.563729887710487, ymax = 11.331837794088093; /* image dimensions */ draw((-8.09002486298222,6.562406202162971)--(-7.673302431346966,0)--(0,0)--cycle, linewidth(0.7)); draw(arc((-8.09002486298222,6.562406202162971),1.5286639717708699,-119.70331682858179,-86.36651254928795)--(-8.09002486298222,6.562406202162971)--cycle, linewidth(0.8)); draw(arc((-4.929590954238111,-8.641330573249373),1.5286639717708699,107.61517948903705,140.95198376833085)--(-4.929590954238111,-8.641330573249373)--cycle, linewidth(0.8)); /* draw figures */ draw((-8.09002486298222,6.562406202162971)--(-7.673302431346966,0), linewidth(0.7)); draw((-7.673302431346966,0)--(0,0), linewidth(0.7)); draw((0,0)--(-8.09002486298222,6.562406202162971), linewidth(0.7)); draw((-7.673302431346966,0)--(-4.929590954238111,-8.641330573249373), linewidth(0.7)); draw((0,0)--(-4.929590954238111,-8.641330573249373), linewidth(0.7)); draw((-8.09002486298222,6.562406202162971)--(-4.929590954238111,-8.641330573249373), linewidth(0.7)); draw((-2.119504865661108,1.719284193992176)--(-2.1195048656611077,-3.715387821385446), linewidth(0.7)); draw((-13.01961581722033,-2.0789243710864023)--(-8.09002486298222,6.562406202162971), linewidth(0.7)); draw((-13.01961581722033,-2.0789243710864023)--(-4.929590954238111,-8.641330573249373), linewidth(0.7)); draw((-13.01961581722033,-2.0789243710864023)--(-7.673302431346966,0), linewidth(0.7)); draw((-13.01961581722033,-2.0789243710864023)--(-6.509807908610165,-1.0394621855432011), linewidth(0.7)); draw((-6.509807908610165,-1.0394621855432011)--(0,0), linewidth(0.7)); draw((-10.900110951559222,1.636463450299044)--(-10.900110951559222,-3.7982085650785784), linewidth(0.7)); draw((-10.346459124283648,-1.0394621855432011)--(-6.509807908610165,-1.0394621855432011), linewidth(0.7) + linetype("2 2")); /* dots and labels */ dot((-8.09002486298222,6.562406202162971),linewidth(2pt) + dotstyle); label("$A$", (-8.273464539594723,6.898712275952563), NE * labelscalefactor); dot((-7.673302431346966,0),linewidth(2pt) + dotstyle); label("$D$", (-8.151171421853054,0.14201752072530618), NE * labelscalefactor); dot((0,0),linewidth(2pt) + dotstyle); label("$B$", (0.34820026119298375,-0.13314199419345088), NE * labelscalefactor); dot((-7.673302431346966,0),linewidth(2pt) + dotstyle); dot((0,0),linewidth(2pt) + dotstyle); dot((-4.929590954238111,-8.641330573249373),linewidth(2pt) + dotstyle); label("$C$", (-5.12441675774673,-9.243979265947852), NE * labelscalefactor); dot((-6.509807908610165,-1.0394621855432011),linewidth(2pt) + dotstyle); label("$M$", (-6.316774655728009,-1.4783662893518188), NE * labelscalefactor); dot((-2.119504865661108,1.719284193992176),linewidth(2pt) + dotstyle); label("$X$", (-2.0059422553341557,1.8541211691086834), NE * labelscalefactor); dot((-2.1195048656611077,-3.715387821385446),linewidth(2pt) + dotstyle); label("$Y$", (-2.0059422553341557,-4.229961438539389), NE * labelscalefactor); dot((-13.01961581722033,-2.0789243710864023),linewidth(2pt) + dotstyle); label("$E$", (-13.59321516135735,-2.4261379518497597), NE * labelscalefactor); label("$\alpha$", (-8.976649966609322,4.544569759425419), NE * labelscalefactor); label("$\alpha$", (-6.408494494034262,-7.256716102645717), NE * labelscalefactor); dot((-10.900110951559222,-3.7982085650785784),linewidth(2pt) + dotstyle); label("$X'$", (-11.48365888031355,-4.352254556281059), NE * labelscalefactor); dot((-10.900110951559222,1.636463450299044),linewidth(2pt) + dotstyle); label("$Y'$", (-11.605951998055218,1.6095349336253437), NE * labelscalefactor); dot((-10.346459124283648,-1.0394621855432011),linewidth(2pt) + dotstyle); label("$N$", (-10.566460497251027,-0.8363274212080523), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Lemma: Let be $D$ a point inside the triangle $AEC$ such that $\angle EAD=\angle ECD=\alpha.$ Let be $M$ the midpoint of the side $AC$ and $Z,W$ the feet of $D$ on the sides $EC$ and $EA$, respectively. Then, $$MZ=MW\,\, \text{and} \,\,\angle ZMW=2\alpha.$$Proof: Let be $J$ and $L$ the midpoints of the segments $CD$ and $AD$, respectively. We have $ZJ=CJ=JD=LM$ and $WL=AL=DL=MJ$. Furthermore, $\angle DJZ=2\alpha=\angle DLW$. So, $\Delta ZJM\equiv \Delta MLW$, such that $MZ=MW$. Also, we get $\angle JZM=\angle LMW=x$ and $\angle JMZ=\angle LWM=y$. But $LMJD$ is a parallelogram, such that $\angle MLD=\angle DJM=z$. In triangle $ZJM$, $x+y+z+2\alpha=180^{\circ}$ and in the parallelogram $JMLD$, $$\angle JML + \angle MLD = 180^{\circ} \therefore x+y+z+\angle ZMW=180^{\circ}.$$[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -17.880027679686517, xmax = 4.035323769514384, ymin = -12.42710343259878, ymax = 10.292973284014675; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-8.142851943995051,4.458715095273911)--(-9.188994791631325,0), linewidth(0.7)); draw(arc((-8.142851943995051,4.458715095273911),1.3412087790208629,-132.0563007819495,-103.20439288131678)--(-8.142851943995051,4.458715095273911)--cycle, linewidth(0.8)); draw(arc((-5.390586110021129,-5.975034710413329),1.3412087790208629,122.44466063474592,151.2965685353786)--(-5.390586110021129,-5.975034710413329)--cycle, linewidth(0.8)); 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label("$W$", (-11.308104662484288,1.5751162203790512), NE * labelscalefactor); dot((-10.829868545642926,-2.996691143568798),linewidth(2pt) + dotstyle); label("$Z$", (-11.200807960162619,-3.5751254910610712), NE * labelscalefactor); dot((-7.289790450826227,-2.9875173552066645),linewidth(2pt) + dotstyle); label("$J$", (-7.0967090963587784,-3.0922903306135594), NE * labelscalefactor); dot((-8.665923367813189,2.2293575476369556),linewidth(2pt) + dotstyle); label("$L$", (-8.518390402120893,2.165248083148232), NE * labelscalefactor); label("$2\alpha$", (-8.786632157925066,-2.609455170166048), NE * labelscalefactor); label("$2\alpha$", (-9.83277500556134,1.280050288994461), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Then, we conclude that $\angle ZMW=2\alpha$. From the lemma above we claim that $X'=Z$ and $Y'=W$ satisfy the problem. It lasts to prove that $ZW\bot MN$ ($N$ is the midpoint of $DE$). But, $DWEZ$ is cyclic, with diameter $AD$. As $NZ=NW$ and $MZ=MW$, we conclude that $MN$ is the perpendicular bisector of the segment $ZW$, and then $ZW\bot MN$, and we are done. This lemma shown above is a well known result and was proposed, maybe the first time, at Australian Mathematical Olympiad, 1983.