The point $X$ is marked inside the triangle $ABC$. The circumcircles of the triangles $AXB$ and $AXC$ intersect the side $BC$ again at $D$ and $E$ respectively. The line $DX$ intersects the side $AC$ at $K$, and the line $EX$ intersects the side $AB$ at $L$. Prove that $LK\parallel BC$.