$ABC$ is a triangle and $DAE$ is a straight line parallel to $BC$ such that $\overline{DA} = \overline{AE}$. If $CD$ meets $AB$ at $X$, and $BE$ meets $AC$ at $Y$, prove that $XY$ is parallel to $BC$.
I was trying to use the Intercept theorem or basic proportionality theorem but I can't see how to use the fact that $\overline{DA}=\overline{AE}$? Any good hint would be sufficient.
Since $DE$ is parallel to $CB$ you will get that similarity between the triangles $\Delta DAX\sim \Delta BCX$ and $\Delta AEY \sim \Delta CBY$.
Since the triangles are similar we have a multiplication factor. This implies that $$\frac{AC}{AY}=1+\frac{CY}{AY}=1+\frac{CB}{AE}=1+\frac{CB}{AD}=1+\frac{BX}{AX}=\frac{AB}{AX}$$ This together with $\angle CAB=\angle YAX$ implies that $\Delta AYX$ is similar to $\Delta ACB$ and will result in the fact that $YX$ is parallel to $CB$