## In the figure below ABCE is a paralleogram. F and G are an points on CE and EB respectively such that FG is parallel to CB. Prove that area of $\Delta A C F = \Delta A B G$

May 21, 2018

see explanation.

#### Explanation:

let $| A B C |$ denote area of $A B C$,
let $B G = w , G E = x , C F = y , F E = z$,
given $F G$ // $C B$,
$\implies \frac{w}{w + x} = \frac{y}{y + z}$
$| A B G \frac{|}{|} A B E | = \frac{w}{w + x}$
$| A C F \frac{|}{|} A C E | = \frac{y}{y + z}$
As $| A B E | = | A C E | = \frac{1}{2} \cdot | A B E C |$
=> |ABG|:|ACF|=w/(w+x):y/(y+z)=1:1
$\implies | A C F | = | A B G |$