$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC. AE$ intersects $CD$ at $F$. Prove that $ar(\triangle ADF) = ar(\triangle ECF)$.
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 prove that $ar(\triangle ADF) = ar(\triangle ECF)$.
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,
$\operatorname{ar}(\Delta \mathrm{ADF})=a r(\Delta \mathrm{CEF})$
Hence proved.
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