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$ABCD$ 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 $ \operatorname{ar}(\mathrm{ADEG})=\operatorname{ar}(\mathrm{GBCE}) $.
Given:
$ABCD$ 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 $BF = 2FC$.
To do:
We have to prove that \( \operatorname{ar}(\mathrm{ADEG})=\operatorname{ar}(\mathrm{GBCE}) \).
Solution:
Draw $EP \perp AB$ and $EQ \perp BC$
$A B=2 G B, C E=2 D E$ and $\mathrm{BF}=2 \mathrm{FC}$
This implies,
$A B-G B=2 G B$
$C D-D E=2 D E$
$\mathrm{BC}-\mathrm{FC}=2 \mathrm{FC}$
$\mathrm{AB}=3 \mathrm{BG}$ and $\mathrm{CD}=3 \mathrm{DE}$
$\mathrm{BC}=3 \mathrm{FC}$
$\mathrm{GB}=\frac{1}{3} \mathrm{AB}$, $\mathrm{DE}=\frac{1}{3} \mathrm{CD}$ and $\mathrm{FC}=\frac{1}{3} \mathrm{BC}$.........(i)
$\operatorname{ar}(\mathrm{ADEG})=\frac{1}{2}(\mathrm{AG}+\mathrm{DE}) \times \mathrm{EP}$
$\operatorname{ar}(\mathrm{ADEG})=\frac{1}{2}(\frac{2}{3}\mathrm{AB}+\frac{1}{3} \mathrm{CD}) \times \mathrm{EP}$
$\Rightarrow \operatorname{ar}(\mathrm{ADEG})=\frac{1}{2}(\frac{2}{3} \mathrm{AB}+\frac{1}{3} \mathrm{AB}) \times \mathrm{EP}$
$\Rightarrow \operatorname{ar}(\mathrm{ADEG})=\frac{1}{2} \times \mathrm{AB} \times \mathrm{EP}$.........(ii)
$a r(\mathrm{GBCE})=\frac{1}{2}(\mathrm{GB}+\mathrm{CE}) \times \mathrm{EP}$
$\Rightarrow \operatorname{ar}(\mathrm{GBCE})=\frac{1}{2}[\frac{1}{3} \mathrm{AB}+\frac{2}{3} \mathrm{CD}] \times \mathrm{EP}$
$\Rightarrow \operatorname{ar}(\mathrm{GBCE})=\frac{1}{2}[\frac{1}{3} \mathrm{AB}+\frac{2}{3} \mathrm{AB}] \times \mathrm{EP}$
$\Rightarrow \operatorname{ar}(\mathrm{GBCE})=\frac{1}{2} \times \mathrm{AB} \times \mathrm{EP}$............(iii)
From (ii) and (iii), we get,
$\operatorname{ar}(\mathrm{ADEG})=\operatorname{ar}(\mathrm{GBCE})$
Hence proved.
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