Triangles $\Delta BCD$ and $\Delta ACE$ are erected externally to a triangle $\Delta ABC$ such that $AE = BD$ and $∠BDC +∠AEC = 180^{\circ}$. Let $F$ be a point on the segment $AB$ such that $\frac{AF}{FB} =\frac{CD}{CE}$.
Prove that $$\frac{DE}{CD+CE} =\frac{EF}{BC}=\frac{FD}{AC}$$