Estonian Mathematical Olympiad

We will use directed angles (Fig. 3 and 4 depict both possible configurations). As $DE$ is a midline in $ABC$, we have $DE\parallel AB$. Thus $$\angle DEY=\angle DEB=\angle ABE=\angle ABY=\angle AXY=\angle DXY.$$ Therefore points $D$, $E$, $X$, $Y$ are concyclic. This means that the circumcircle of $EXY$ intersects $DE$ at $D$. Thus $K=D$, i.e. the midpoint of $BC$, regardless of the choice of $X$.