36. Exemplar exercise problems IX

In the given figure, line segment $DF$ intersect the side $AC$ of a triangle $ABC$ at the point $E$ such that $E$ is the mid-point of $CA$ and $\angle AEF = \angle AFE$. Prove that $\frac{BD}{CD} = \frac{BF}{CE}$. [Hint: Take point $G$ on $AB$ such that $CG||DF$.]