Question

$ABCD$ is a parallelogram in which $BC$ is proved to $E$ such that $CE=BC$. $AE$ intersect $CD$ at $F$.
Prove that If the area of $△DFB=3c{m}^2$, find the area (in $c{m}^2$) of $|{|}^{gm}ABCD$.

Answer 1

In $△ADF$ and $△ECF$,

$\Rightarrow$ $\angle ADF=\angle ECF$ [ Alternate interior angles, since $AD\parallel CE$ ]

$\Rightarrow$ $AD=EC$ [ Since, $AD=BC=CE$ ]

$\Rightarrow$ $\angle DFA=\angle CFA$ [ Vertically opposite angles ]

$\Rightarrow$ $△ADF\cong △ECF$ [ By $AAS$ congruence criteria ]

$\Rightarrow$ $ar\left(△ADF\right)=ar\left(△ECF\right)$

$\Rightarrow$ $DF=CF$ [ C.P.C.T. ]

Now,

$DF=CF$

$\Rightarrow$ $BF$ is a median in $△BCD$

$\Rightarrow$ $ar\left(△BCD\right)=2\times ar\left(△BDF\right)$

$\Rightarrow$ $ar\left(△BCD\right)=2\times 3c{m}^2$

$\Rightarrow$ $ar\left(△BCD\right)=6c{m}^2$

Now,

Area of parallelogram $ABCD=2\times ar\left(△BCD\right)$

$=2\times 6c{m}^2$

$=12c{m}^2$