Question

$ABCD$ is a parallelogram, $G$ is the point on $AB$ such that $AG=2GB,E$ is a point of $DC$ such that $CE=2DE$ and F is the point of BC such that $BF=2FC$. Prove
that: $ar\left(ADEG\right)=ar\left(GBCE\right)$

Answer 1

Let $ABCD$ be a rectangle (a type of parallelogram)

$AG=2GB$

$CE=2DE$

$BF=2FC$

a) $as\left(ADEG\right)=as\left(GBCE\right)$

$\Rightarrow CD=AB=lm$

In $ADEG$(trapezium)

$\therefore AD=b$

$AG=\frac{2}{3}$ and $DE=\frac{1}{3}l$

$\therefore asADEG=\frac{1}{2}\times [sumof||sides\times Distanceb(w\left)\right]$

$=\frac{1}{2}\times (\frac{2}{3}l+\frac{1}{3}l)\times b$

$=\frac{1}{2}\times l\times b$

In $GBCE$

$BC=b$

$GB=\frac{1}{3}l,EC=\frac{2}{3}l$

as $GBCE=\frac{1}{2}\times (\frac{1}{3}l+\frac{2}{3}l)\times b$

$=\frac{1}{2}\times l\times b$

as $ADGE=asGBCE,H.P$

b) $△EGB=\frac{1}{6}\left(ABCD\right)$

$as\left(ABCD\right)=l\times b(\therefore \Rightarrow AB\times BC)$

$ar\left(EGB\right)=ar\left(EGB\right)-ar\left(EXB\right)-a\left(EXG\right)$

$=\frac{1}{2}\times \left(EX\right)\times \left(XB\right)-\frac{1}{2}\times X\left(EX\right)\times \left(XG\right)$

$\frac{1}{2}\times b\times \frac{2}{3}l-\frac{1}{2}\times b\times \frac{1}{3}l$

$=\frac{1}{2}\times b\times l(\frac{2}{3}-\frac{l}{3})$

$ar\left(△EGB\right)=\frac{1}{2}\times \frac{1}{3}\times b\times l=\frac{1}{6}\times ar\left(ABCD\right)$hence prove

C) $ar\left(△EFG\right)=\frac{1}{2}ar\left(△EFG\right)$

$\frac{1}{2}\times \left(EC\right)\times \left(CF\right)=\frac{1}{2}\left[ar\right(△EFG)-ar(△EFC\left)\right]$

$(1+\frac{1}{2})(\frac{1}{2}\times (EC)\times (FC\left)\right)$ $=\frac{1}{2}ar\left(\frac{1}{2}\right(EC)\times BC)$

$\frac{3}{2}\times \frac{1}{2}\times \frac{2}{3}l\times \frac{1}{3b}$ $=\frac{1}{2}\times \frac{1}{2}\times \frac{2}{3}l\times b$

$\frac{2}{2\times 2\times 2\times 3}lb$ $=\frac{2}{2\times 2\times 3}lb$

d) as $\left(△EGB\right)=\frac{1}{2}ar\left(△EFG\right)$

$\frac{1}{6}ar\left(ABCD\right)=\frac{1}{2}(\frac{1}{2}\times \frac{2}{9}ar(ABCD))$

$\frac{1}{6}=\frac{1}{2}\times \frac{1}{3}\times \frac{1}{3}$

$\therefore$ Not True

$ar\frac{2}{3}ar\left(△EGB\right)=ar\left(△EFC\right)$

$or\frac{1}{3}ar\left(△EGB\right)=\frac{1}{2}ar\left(△EFG\right)$

e) $ar△EFG=ar\left(△EGB\right)+\left(△BFE\right)=(\frac{1}{6}+\frac{2}{9})ar\left(ABCD\right)$

$=\left(\frac{7}{18}\right)ar\left(ABCd\right)$