In figure , $A B C D E$ is any pentagon . $B P$ drawn parallel to $A C$ meets $D C$ produced at $P$ and $E Q$ drawn parallel to $A D$ meets $C D$ produced at $Q$ .
Prove that $a r (A B C D E) = a r (A P Q)$ .

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Solution

$B P | | A C$ and $A D | | E Q$

Since, triangles on the same base and between the same parallels are equal in area

$a r (Δ A B C) = a r (Δ A P C)$ .........(1)

and $a r (Δ A D E) = a r (Δ A D Q)$ .........(2)

Adding (1) and (2), we get

$a r (Δ A B C) + a r (Δ A D E)$ $= a r (Δ A P C) = + a r (Δ A D Q)$

Adding $a r (Δ A C D)$ to both sides, we get

$a r (Δ A B C) + a r (Δ A D E) + a r (Δ A C D)$ $= a r (Δ A P C) + a r (Δ A D Q) + a r (Δ A C D)$

Hence, $a r (A B C D E) = a r (Δ A P Q)$

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