In $\triangle ABC$, $XY$ is drawn parallel to $BC$ cutting $AB$ and $AC$ at $X$ and $Y$. Prove that $BY$ and $CX$ intersect in the median through $A$.
I've constructed the median $AD$ which intersects $XY$ at $E$. Let $BY$ and $CX$ intersect at $O$.
I don't know if it's of any use but $\triangle YEO$ is similar to $\triangle BDO$ and $\triangle XEO$ is similar to $\triangle CDO$. I've just considered the trapezium $XYCB$ and am trying to prove that the line joining the midpoints of the parallel sides passes through the intersection of the diagonals.
I am not able to proceed further. Please help.