Question

ABCD is a parallelogram, G is the point on AB such that $AG=2GB$. E is a point on DC such that $CE=2DE$ and F is a point of BC such that $BF=2FC$. Prove that
$ar\left(△EFC\right)=\frac{1}{2}ar\left(△EBF\right)$

Answer 1

Given$ABCD$ is a parallelogram in which $AG=2GB,CE=2DE$ and $BF=2FC$

$1)$ Since $ABCD$ is a parallelogram we have $AB\parallel CD$ and $AB=CD$

$\therefore BG=\frac{1}{3}AB$ and $DE=\frac{1}{3}AB$

$\therefore BG=DE$

$\therefore ADEH$ is a parallelogram (since $AH\parallel DE$ and $AD\parallel HE$)

Area of parallelogram $ADEH$=area of parallelogram $BCIG$ ....$\left(1\right)$

Since $DE=BG$ and $AD=BC$ parallelogram with corresponding sides are equal

we have area$\left(△HEG\right)=$area$\left(△EGI\right)$ ....$\left(2\right)$

Diagonals of a parallelogram divide it into two equal areas.

From $\left(1\right)$ and $\left(2\right)$ we get

Area of parallelogram $ADEH+$area$\left(△HEG\right)=$area of parallelogram $BCIG+$area$\left(△EGI\right)$

$\therefore$ Area of parallelogram $ADEG=$Area of parallelogram $GBCE$

$2)$Height of parallelogram $ABCD$ and $△EGB$ is the same

Base of $△EGB=\frac{1}{3}AB$

Area of parallelogram $ABCD=h\times AB$ area$\left(△EGB\right)=\frac{1}{2}\times h\times \frac{1}{3}AB=\frac{1}{6}h\times AB$

Area$\left(△EGB\right)=\frac{1}{6}$Area of parallelogram $ABCD$

$3)$Let the distance between $EH$ and $CB=x$

area$\left(△EBF\right)=\frac{1}{2}\times BF\times x=\frac{1}{2}\times \frac{2}{3}BC\times x=\frac{1}{3}\times BC\times x$

and area$\left(△EFC\right)=\frac{1}{2}\times CF\times x=\frac{1}{2}\times \frac{1}{3}BC\times x=\frac{1}{2}$Area of $\left(△EBF\right)$

$\Rightarrow$ Area$\left(△EFC\right)=\frac{1}{2}\times$Area$\left(△EBF\right)$