In the figure,ABCD and AEFD are two parallelo | vedclass

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(N/A) Given: $ABCD$ and $AEFD$ are two parallelograms.To prove: $\operatorname{ar}(	riangle PEA) = \operatorname{ar}(	riangle QFD)$.Proof:In $	rian

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(N/A) Given: $ABCD$ and $AEFD$ are two parallelograms.
To prove: $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.
Proof:
In $\triangle PEA$ and $\triangle QFD$:
$1$. $\angle APE = \angle DQF$ (Corresponding angles are equal as $AB \parallel CD$ and $PQ$ is a transversal).
$2$. $\angle AEP = \angle DFQ$ (Corresponding angles are equal as $AE \parallel DF$ and $PQ$ is a transversal).
$3$. $AE = DF$ (Opposite sides of parallelogram $AEFD$ are equal).
Therefore,by $AAS$ congruence rule,$\triangle PEA \cong \triangle QFD$.
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

In the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

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