Belarusian Mathematical Olympiad

First we prove that the points $A$, $L$, $X$ and $K$ lie on the same circle. Since the quadrilateral $ABDX$ is cyclic, $\angle AXK=\angle ABC$. Similarly $\angle LXA=\angle BCA$. Now from the equality $\angle KAL+\angle LXK=\angle CAB+\angle ABC+\angle BCA=180^{\circ }$ it follows that the quadrilateral $ALXK$ is cyclic. Therefore $\angle KLX=\angle KAX$.

Since the points $A$, $X$, $E$ and $C$ lie on the circle, $\angle DEX=\angle CAX$. So $\angle KLX=\angle DEX$ and $LK\parallel BC$.