Remark. $OXYC$ is cyclic too .

hint

$O$ is the circumcenter of $\triangle ABC,$ $P$ is the foot of the C-altitude and $D,E$ are the midpoints of $BC,CA.$ Since $\triangle AEP$ and $\triangle BDP$ are isosceles with apices $E,D,$ it follows that $PE \parallel MX$ and $PD \parallel MY.$ Thus $\frac{EX}{XA}=\frac{PM}{MA} \ , \ \frac{DY}{YB}=\frac{PM}{MB} \Longrightarrow \frac{EX}{DY}=\frac{XA}{MA} \cdot \frac{MB}{YB}=\frac{EA}{DB} \cdot \frac{PB}{PA}=\frac{OE}{OD}$ Thus, the right $\triangle OEX$ and $\triangle ODY$ are similar. As a result, $\angle XOY=\angle EOD,$ i.e. $C,X,Y,O$ are concyclic. Let $S,T$ be the midpoints of $MA,MB.$ If $\odot(MXY)$ cuts $YT$ again at $U,$ then $\angle OXY=\angle OCY=\angle MXS$ implies that $\angle OXS=\angle MXY=\angle MUT=\angle BUT \Longrightarrow OX \parallel BU$ But $\angle AXU=\angle AXM+\angle MYU=90^{\circ}-\angle A+\angle C \Longrightarrow UX \parallel BO$ Hence, $OBUX$ is a parallelogram with diagonal intersection $L$ $\Longrightarrow$ $U$ coincides with the reflection $O_2$ of $O$ about $L$ $\Longrightarrow$ $O_2 \in \odot(MXY).$ By analogous reasoning, we'll have that $O_1 \in \odot(MXY).$

Call $O_1$=E and $O_2$=D. Look at the lines perpendicular to AB that pass through X, Y, O, K and L. The first two bisect AM and MB, respectively, so call the midpoint of AM F and the midpoint of MB G. The third line bisects AB. The fourth line bisects AG and the fifth line bisects BF. Using this we easily prove that the distance between the 1º and 4º line is the same as the distance between the 4º and 3º lines, and therefore D is on the first line. Analogously, E is on the second line. Now we see OBEX and OADY are parallelograms. Therefore <DXE = 90 - <ABO = 90 - <BAO = <EYD and therefore XDEY is cyclic. The End.

Also, M is also on this circle because <XMY = <BCA = 90 - <ABO = <YEX. So MXYDE is a cyclic pentagon. Another remark, OXYC is cyclic, which is easy to see.

.Let $P$ be the midpoint of $AB \Rightarrow .PL=\frac{1}{2}XA=\frac{1}{2}MX$ .Similarly $PK=\frac{1}{2}MY$.Observe that $\angle YMX=\angle A =\angle LPK \Rightarrow \Delta LPK \sim \Delta XMY \Rightarrow LK=\frac{1}{2}XY \Rightarrow O_1O_2=XY$ .Since $YOAO_1$ and $OXO_2B$ are parallelograms $XO_2=OB=OA=O_1Y$ .Hence $XYO_2O_1$ is a isosceles trapezoid.Hence proved .

my proof also similar to @MMEEnV 's proof! We will prove that $XYO_2O_1$ is isosceles trepazoid. We have that $4KL^2=AX^2+XY^2+BY^2+AB^2-AY^2-BX^2=XY^2+AB^2-AM\cdot AB-BM\cdot AB=XY^2$. So $O_1O_2=XY$ ...