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Question
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O
intersects AB at P and DC at Q. Prove that ar (Δ POA) = ar (Δ QOC).
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Solution
In triangles and , POA QOC we have
∠AOP = ∠COQ [vertically opposite angles]
OA = OC [ Diagonals of a parallelogram bisect each other]
∠PAC = ∠QCA [ AB || DC ; alternative angles ]
So, by ASA congruence criterion, we have
ΔPOA ≅ QOC
Area (ΔPOA) = area (ΔQOC) .
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