# triangles

• Nov 18th 2009, 06:43 AM
thereddevils
triangles
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 $\angle ACD$. Prove that DC=DE
• Nov 18th 2009, 09:46 AM
red_dog
$\Delta FAB\sim\Delta FDE\Rightarrow\frac{AB}{DE}=\frac{AF}{FD}$

In the triangle ACD apply the bisector theorem: $\frac{AF}{FD}=\frac{AC}{CD}$

Then $\frac{AB}{DE}=\frac{AC}{CD}$.

But $AB=AC\Rightarrow DE=CD$