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In the figure, $PSDA$ is a parallelogram in which $PQ = QR = RS$ and $AP \parallel BQ \parallel CR$. Prove that $ar(\triangle PQE) = ar(\triangle CFD)$.
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Academic Mathematics NCERT Class 9

Given:

$PSDA$ is a parallelogram in which $PQ = QR = RS$ and $AP \parallel BQ \parallel CR$.

To do:

We have to prove that $ar(\triangle PQE) = ar(\triangle CFD)$.

Solution:

$PA \parallel BQ \parallel CR \parallel DS$ and $PQ - QR = RS$

Therefore,

$AB = BC = CD$

$PQ = CD$

In $ABED, F$ is the mid point of $ED$

This implies,

$EF = FD$

Similarly,

$EF = PE$

$PE = FD$

In $\triangle PQE$ and $\triangle CFD$,

$\angle EPQ = \angle FDC$ (Alternate angles)

$PQ = CD$

$PE = FD$

Therefore, by SAS axiom,

$\triangle PQE \cong \triangle CFD$

This implies,

$ar(\triangle PQE) = ar(\triangle CFD)$

Hence proved.

Updated on: 10-Oct-2022

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