In the figure, $ABCD$ and $AEFD$ are two parallelograms. Prove that $ar(\triangle APE) : ar(\triangle PFA) = ar(\triangle QFD) : ar(\triangle PFD)$.
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Given:
$ABCD$ and $AEFD$ are two parallelograms.
To do:
We have to prove that $ar(\triangle APE) : ar(\triangle PFA) = ar(\triangle QFD) : ar(\triangle PFD)$.
Solution:
In $\triangle AEP$ and $\triangle DFQ$,
$AE = DF$ (Opposite sides of a parallelogram)
$\angle AEP = \angle DFQ$ (Corresponding angles)
$\angle APE = \angle DQF$ (Corresponding angles)
Therefore, by AAS axiom,
$\triangle AEP \cong \triangle DFQ$
This implies,
$PE = QF$ (CPCT)
$ar(\triangle AEP) = ar(\triangle DFQ)$.........(i)
$\triangle PFA$ and $\triangle PFD$ are on the same base $PF$ and between the same parallels
This implies,
$ar(\triangle PFA) = ar(\triangle PFD)$......…(ii)
From (i) and (ii), we get,
$\frac{\operatorname{ar}(\Delta \mathrm{AEP})}{a r(\Delta \mathrm{PFA})}=\frac{\operatorname{ar}(\Delta \mathrm{DFQ})}{\operatorname{ar}(\Delta \mathrm{PFD})}$
Hence proved.
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