In the adjoining figure, ABCD is a $| |^{g m}$ whose diagonals AC and BD intersect at O. A line segment through O meets AB at P and DC at Q. Prove that ar(APQD) $= \frac{1}{2}$ ar(ABCD).

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Solution

Diagonal AC of $| |^{g m}$ ABCD divides it into two triangles of equal area.
$∴ a r (Δ A C D) = \frac{1}{2} a r (A B C D) \dots . (i)$
$In △ O A P and ∠ O C Q, we have$
$O A = O C [diagonals of a | |^{g m} bisect each other]$
$∠ A O P = ∠ C O Q [vert. opp. angles]$
$∠ P A O = ∠ Q C O [alt. int. angles]$
$∴ △ O A P ≅ △ O C Q [by ASA]$
$\Rightarrow a r (△ O A P) = a r (△ O C Q)$
$\Rightarrow a r (△ O A P) + a r (A O Q D) = a r (△ O C Q) + a r (A O Q D)$
$\Rightarrow a r (A P Q D) = 2 a r (△ A C D) = \frac{1}{2} a r (A B C D) [using (i)]$
$∴ a r (A P Q D) = \frac{1}{2} a r (A B C D)$

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