In the given figure, line-segment $\mathrm{DF}$ intersects the side $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ at point $\mathrm{E}$ such that $\mathrm{E}$ is the midpoint of $\mathrm{CA}$ and $\angle \mathrm{AEF}=\angle \mathrm{AFE}$
Prove that$: \mathrm{BD}:\mathrm{CD}=\mathrm{BF}:\mathrm{CE}$.