Find the value of the expression $\frac{AF}{FD}$.

79 Views Asked by Bumbble Comm At 04 May 2026 - 10:00

In the given figure $AC=BD=3$ units and $CE=DE=1$ unit.

We know that the median divides the area of a triangle in to two equal halves.

Therefore,

AREA $\triangle ECD$ = $\frac{1}{2}$$\frac{1}{2}$(AREA $\triangle EAD$)

enter image description here

There are 1 best solutions below

Bumbble Comm On 03 Jun 2013 - 1:22

Let $r$ be the bisector of $\angle AEB$. By simmetry, $F$ can be defined as $AD\cap r$. So $\frac{AF}{FD}=\frac{AE}{ED}=4$.