Question
In Fig. $ABCD$ and $AEFD$ are two parallelograms. Prove that $ar\ (PEA) = ar\ (QFD).$
✓
Answer
We have to prove that $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})$ join $PD.$
In $\triangle\text{PEA}$ $\triangle\text{QFD}$ we have
$\angle\text{APE}=\angle\text{DQF}[\because$ corresponding $\angle\text{S}$ are equal as $AB || CD]$
$\angle\text{APE}=\angle\text{DFQ}[\because$ corresponding $\angle\text{S}$ are equal as $AE || DF]$
$AE = DF [\because$ opposite sides of $a || gm$ are equal$]$
$\therefore\triangle\text{PEA}\cong\triangle\text{QFD} [$By $AAS$ Cong. Rule$]$
Hence, $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QED})$
In $\triangle\text{PEA}$ $\triangle\text{QFD}$ we have
$\angle\text{APE}=\angle\text{DQF}[\because$ corresponding $\angle\text{S}$ are equal as $AB || CD]$
$\angle\text{APE}=\angle\text{DFQ}[\because$ corresponding $\angle\text{S}$ are equal as $AE || DF]$
$AE = DF [\because$ opposite sides of $a || gm$ are equal$]$
$\therefore\triangle\text{PEA}\cong\triangle\text{QFD} [$By $AAS$ Cong. Rule$]$
Hence, $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QED})$
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