A really nice problem. Let $\omega$ be a circle through $M$ and tangent to the line parallel $\ell$ to $BC$ through $A$. Redefine $X$ and $Y$ as intersection of $\omega$ with $AB$ and $AC$ respectively. (We can see that $AX > BX$ and $AY > CY$). Because $M$ is the midpoint $BC$, then $-1=(B,C;M,\infty_{BC}) = (X,Y;M,A)$, hence $AXMY$ is harmonic, which means that $\angle KXM = \angle AXY = \angle ACB$, last equality is from tangency in $A$ and parallel lines $\ell$ and $BC$. Same with $\angle KYM = \angle AYX = \angle ABC$, hence $X$ and $Y$ are the same as given in the statement. But now by $\angle AXY = \angle ACB$ we have that $B,C,X,Y$ lie on a circle.

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Remark: there is no need that both $AX>BX, AY>CY, X, Y$ must not be the midpoints of the sides, unless $AB=AC$. From the wording I guess you got the idea of solution! Best regards, sunken rock

Hmm X is the Second intersection of circle KMN with AB where N is the midpoint of AB. Similarly define Y. From PoP we get AX = AM^2/AB and AY = AM^2/AC, so we have AX*AB = AY*AC hence done.

Very cute, let $P=(A+B)/2$ and $Q=(A+C)/2$, note that $P,K,Q$ are collinear and $PQMB$ and $QPMC$ are parallelograms so $\measuredangle KQM=\measuredangle KYM$ and $\measuredangle KPM = \measuredangle KXM$ which means \[AX \cdot AB = 2AX \cdot AP = 2AK \cdot AM = 2 AY\cdot AQ = AY \cdot AC ~ \blacksquare\]

Let $D$ and $E$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, so $DKMX$ and $EKMY$ are cyclic. Therefore, $AD\cdot AX=AK\cdot AM=AE\cdot AY \implies AB\cdot AX=AC\cdot AY$; done. $\blacksquare$