Question

Please prove?

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 `DeltaACF=DeltaABG`

Answer 1

see explanation.

Explanation:

let `|ABC|` denote area of `ABC`,
let `BG=w, GE=x, CF=y, FE=z`,
given `FG` // `CB`,
`=> w/(w+x)=y/(y+z)`
`|ABG|/|ABE|=w/(w+x)`
`|ACF|/|ACE|=y/(y+z)`
As `|ABE|=|ACE|=1/2*|ABEC|`
`=> |ABG|:|ACF|=w/(w+x):y/(y+z)=1:1`
`=> |ACF|=|ABG|`