Question 7
In the figure, ABCDE is any pentagon, BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar ( $Δ$ APQ).

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Solution

We know that triangles on the same base and between the same parallels are equal in area.
Here, $Δ A D Q$ and $Δ A D E$ lie on the same base AD and between the same parallels AD and EQ.
$a r (Δ A D Q) = a r (Δ A D E)$ ....(i)
$a r (Δ A C P) = a r (Δ A C B)$ ....(ii)
On adding eqs. (i) and (ii) we get,
$a r (Δ A D Q) + (Δ A C P) = a r (Δ A D E) + a r (Δ A C B)$
On adding $a r (Δ A C D)$ both sides, we get,
$a r (Δ A D Q) + a r (Δ A C P) + a r (Δ A C D) = a r (Δ A D E) + a r (Δ A C B) + a r (Δ A C D)$
$\Rightarrow a r (Δ A P Q) = a r (A B C D E)$

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