Triangle $ABC$ $XY$ parallel to $BC$

Question

In a triangle $ABC$ $XY$ is drawn parallel to $BC$ cutting $AB$ and $AC$ in $X$ and $Y$, respectively.Prove that $BY$ and $CX$ intersect on the median through $A$. I have tried using Menelaus theorem on $\triangle ABO$ where $O$ is the intersection of the median through $A$ and the line $BC$ and $BC$ is the extended line

Answer 1

Actually this problem is immediately solved by Ceva's Theorem and Thales. Let $E=CX \cap BY$ and $F= AE \cap BC$. Then $$\frac{AX\cdot BF\cdot CY}{XB\cdot FC\cdot YA}=1$$ Also due $BC\parallel XY$ we have $$\frac{AX}{XB}=\frac{AY}{YC}$$ This implies $BF=FC$.