Geometry-Triangle Let $ABC$ be a triangle with $DAE$, a straight line parallel to BC such that $DA=AE$. If $CD$ meets $AB$ at X and $BE$ meets $AC$ at $Y$, prove that $XY$ is parallel to $BC$
I tried to use the angle approach by couldn't work out the problem as I couldn't effectively prove the thing using alternate angles approach.


We have that $\triangle{AXD}$ and $\triangle{BXC}$ are similar and that $\triangle{EYA}$ and $\triangle{BYC}$ are similar, and so
$$AX:BX=AD:BC=AE:BC=AY:CY$$
from which the claim follows.
