# 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 $A$.

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.

This follows immediatelly from the Ceva's theorem, as $\frac{AX}{XB} = \frac{AY}{YC}$. This implies that the line passing through $A$ and the intersection of $CX$ and $BY$ bisects $BC$.