Question

$ABCD$ is a parallelogram. $E$ is a point on $BA$ such that $BE=2EA$ 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 parallelogram $ABCD$.

Answer 1

We have,
$BE=2EA$ and $DF=2FC$

$\Rightarrow AB-AE=2AE$ and $DC-FC=2FC$

$\Rightarrow AB=3AE$ and $DC=3FC$

$\Rightarrow AE=\frac{1}{3}AB$ and $FC=\frac{1}{3}DC$

$\Rightarrow AE=FC[\because AB=DC]$

$\therefore AEFC$ is a parallelogoram.

Clearly, parallelogram $ABCD$ and $AEFC$ have the same altitude and $AE=\frac{1}{3}AB.$

$ar\left(|{|}^{gm}AECF\right)=h\cdot \frac{1}{3}AB=\frac{1}{3}h\cdot AB=\frac{1}{3}ar\left(|{|}^{gm}ABCD\right)$

$\therefore ar\left(|{|}^{gm}AECF\right)=\frac{1}{3}ar\left(|{|}^{gm}ABCD\right)$