In the figure , AFD and BFE are straight lines and AB is parallel to ED . THe point C is such that AB=AC and FC bisects $\displaystyle \angle ACD $. Prove that DC=DE
$\displaystyle \Delta FAB\sim\Delta FDE\Rightarrow\frac{AB}{DE}=\frac{AF}{FD}$
In the triangle ACD apply the bisector theorem: $\displaystyle \frac{AF}{FD}=\frac{AC}{CD}$
Then $\displaystyle \frac{AB}{DE}=\frac{AC}{CD}$.
But $\displaystyle AB=AC\Rightarrow DE=CD$