62nd Ukrainian National Mathematical Olympiad

Let $N$ be the middle of $AB$, and $L$ be the middle of $AC$. Then the points $N$, $K$ and $L$ lie on the same line—the midline of $△ABC$ which is parallel to $BC$ (fig. 19).

Then we have that $\angle KXM=\angle ACB=\angle KNM$, so the quadrilateral $KNXM$ is inscribed, and similarly $KMYL$ is also inscribed.

Therefore, $AN\cdot AX=AK\cdot AM=AL\cdot AY$, so the points $Y$, $N$, $L$ and $X$ lie on the same circle, hence $\angle AXY=\angle ALN=\angle ACB$, so the quadrilateral $BXYC$ is inscribed, which is what we needed to prove.

Remarks: In this configuration, the quadrilateral $AXMY$ is harmonic.