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$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC. AE$ intersects $CD$ at $F$. If the area of $\triangle DFB = 3\ cm^2$, find the area of parallelogram $ABCD$.
Given:
$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC. AE$ intersects $CD$ at $F$.
To do:
We have to find the area of parallelogram $ABCD$.
Solution:
In $\triangle \mathrm{ADF}$ and $\triangle \mathrm{ECF}$,
$\mathrm{AD}=\mathrm{CE}$
$\angle \mathrm{AFD}=\angle \mathrm{CFE}$ (vertically opposite angles)
Therefore, by AAS axiom,
$\triangle \mathrm{ADF} \cong \Delta \mathrm{ECF}$
This implies,
$\operatorname{ar}(\Delta \mathrm{ADF})=a r(\Delta \mathrm{CEF})$
$\mathrm{AF}=\mathrm{CF}$ (CPCT)
$\mathrm{AF}=\mathrm{EF}$ (CPCT)
$\mathrm{BF}$ is the median of $\triangle \mathrm{BCD}$.
This implies,
$\operatorname{ar}(\Delta \mathrm{BFD})=\operatorname{ar}(\Delta \mathrm{BFC})$
$=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{BCD})$
$=\frac{1}{2}[\frac{1}{2} a r(parallelogram \mathrm{gm} \mathrm{ABCD})]$ ($\mathrm{BD}$ is the diagonal of parallelogram)
$=\frac{1}{4} a r(parallelogram \mathrm{ABCD})$
$operatorname{ar}(\triangle \mathrm{BFD})=3 \mathrm{~cm}^{2}$
Therefore,
Area of parallelogram $\mathrm{ABCD}=4 \times \operatorname{ar}(\Delta \mathrm{BFD})$
$=4 \times 3$
$=12 \mathrm{~cm}^{2}$.
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