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$ABCD$ 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 $ABCD$.
Given:
$ABCD$ is a parallelogram. $E$ is a point on $BA$ such that $BE = 2EA$ and $F$ is a point on $DC$ such that $DF = 2FC$.
To do:
We have to prove that $AECF$ is a parallelogram whose area is one-third of the area of parallelogram $ABCD$.
Solution:
Join $AE$ and $CE$.
In parallelogram $\mathrm{ABCD}$,
$\mathrm{AE}=2 \mathrm{EB}$ and $\mathrm{DF}=2 \mathrm{FC}$
$\Rightarrow \mathrm{AE}=\frac{1}{3} \mathrm{AB}$
$\mathrm{CF}=\frac{1}{3} \mathrm{CD}$
$\mathrm{AB}=\mathrm{CD}$ (Opposite sides of a parallelogram)
This implies,
$\mathrm{AE}=\mathrm{FC}$
$\mathrm{AB} \| \mathrm{CD}$
Therefore, $AECF$ is a parallelogram.
Parallelogram $\mathrm{ABCD}$ and parallelogram $AECF$ have the same altitude and $\mathrm{AE}=\frac{1}{3} \mathrm{AB}$
Area of parallelogram $\mathrm{AECF})=\mathrm{AE} \times \text { Altitude }$
$=\frac{1}{3} \mathrm{AB} \times \text { Altitude }$
$=\frac{1}{3} a r(parallelogram \mathrm{ABCD})$.
Hence proved.
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