Question
$A B C D$ is a parallelogram. $E$ is a point on $B A$ such that $B E=2 E A$ and $F$ is point on $D C$ such that $D F=2 F C$. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram $A B C D$.
✓
Answer
Draw $FG \perp AB$
We have, $B E=2 E A$ and $D F=2 F C$
$\Rightarrow A B-A E=2 A E$ and $D C-F C=2 F C$
$\Rightarrow A B=3 A E$ and $D C=3 F C$
$\Rightarrow AE=\left(\frac{1}{3}\right) AB \text { and } FC=\left(\frac{1}{3}\right) DC \ldots$
But $A B=D C$ Then, $A E=F C$ [opposite sides of $\left.\|^{g m}\right]$ Thus, $A E=F C$ and $A E \| F C$ Then, $A E C F$ is a parallelogram Now, area of parallelogram $AECF = AE \times FG \Rightarrow \operatorname{ar}\left(| |^{ gm } AECF \right)=\frac{1}{3} AB \times FG$ from (1)
$\Rightarrow 3 a r\left(\|{ }^{g m} A E C F\right)=A B \times F G \ldots$...(2) And $\operatorname{ar}(\| g gm A B C D)=A B \times F G \cdots(3)$
Compare equation 2 and 3
$\Rightarrow 3 \operatorname{ar}\left(\|^{gm} AECF\right)=\operatorname{ar}\left(\| \|^{gm} ABCD\right)$
$\Rightarrow \operatorname{ar}\left(\|^{gm} AECF\right)=\frac{1}{3} \operatorname{ar}\left(\|^{gm} ABCD\right)$
We have, $B E=2 E A$ and $D F=2 F C$
$\Rightarrow A B-A E=2 A E$ and $D C-F C=2 F C$
$\Rightarrow A B=3 A E$ and $D C=3 F C$
$\Rightarrow AE=\left(\frac{1}{3}\right) AB \text { and } FC=\left(\frac{1}{3}\right) DC \ldots$
But $A B=D C$ Then, $A E=F C$ [opposite sides of $\left.\|^{g m}\right]$ Thus, $A E=F C$ and $A E \| F C$ Then, $A E C F$ is a parallelogram Now, area of parallelogram $AECF = AE \times FG \Rightarrow \operatorname{ar}\left(| |^{ gm } AECF \right)=\frac{1}{3} AB \times FG$ from (1)
$\Rightarrow 3 a r\left(\|{ }^{g m} A E C F\right)=A B \times F G \ldots$...(2) And $\operatorname{ar}(\| g gm A B C D)=A B \times F G \cdots(3)$
Compare equation 2 and 3
$\Rightarrow 3 \operatorname{ar}\left(\|^{gm} AECF\right)=\operatorname{ar}\left(\| \|^{gm} ABCD\right)$
$\Rightarrow \operatorname{ar}\left(\|^{gm} AECF\right)=\frac{1}{3} \operatorname{ar}\left(\|^{gm} ABCD\right)$
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