Question

ABCD is a parallelogram. E is a point on BA such That BE = 2 EA and F is a point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of paralllelogram ABCD.

Answer 1

Given: In ||gm ABCD, E is a point on AB such that BE = 2EA and F is a point on CD such that DF = 2FC. AE and CE are joined

To prove : AECF is a || gm and
ar $\left(||gmAECF\right)=\frac{1}{3}ar\left(||gmABCD\right)$
Proof : (i) in || gm ABCD,
AE= 2 EB and DF = 2 FC
$\Rightarrow AE=\frac{1}{3}ABandCF=\frac{1}{3}CD$
But AB = CD
\therefore AE = FC
and AB || CD
\therefore AECF is a ||gm
(ii) Clearly || gm ABCD and ||gm AECF have the Same altitude and $AE=\frac{1}{3}AB$
$\therefore Area\left(||gmAECF\right)=base\times altitude=AE\times altitude=\frac{1}{3}AB\times altitude=\frac{1}{3}ar\left(||gmABCD\right)$