In fig, $A B C D$ is a parallelogram and $B C$ is produced to a point $Q$ such that $A D = C Q$ . If $A Q$ intersect $D C$ at $P$ , show that $a r (B P C) = a r (D P Q)$ .

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Solution

Join $A C$ .
As we know that area of triangles on the same base and between the same parallels lines are equal
Therefore , $a r (A P C) = a r (B P C)$ ... (1)
Now In quadrilateral $A C Q D$ , we have
$A D = C Q$
and , $A D ∥ C Q$ [Given]

Therefore, this quadrilateral $A D Q C$ with one pair of opposite sides is equal and parallel is parallelogram.
Therefore $A D Q C$ is a parallelogram.
$\Rightarrow A P = P Q$ and $C P = D P$
[Since diagonals of a|| gm bisect each other]
In $Δ s A P C$ and $D P Q$ we have
$A P = P Q$ [Proved above]
$∠ A P C = ∠ D P Q$ [Vertically opp. $∠ s$ ]
and ,
$P C = P D$ [Proved above]
Therefore by SAS criterion of congruence,
$Δ A P C ≅ Δ D P Q$

$\Rightarrow a r (A P C) = a r (D P Q)$ ... (2)
[Since congruent $Δ s$ have equal area]
Therefore $a r (B P C) = a r (D P Q)$ [From (1)]
Hence , $a r (B P C) = a r (D P Q)$

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