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Question
P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
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Solution
Given: ABCD is a parallelogram whose diagonals bisect each other at O.
To show: PQ is bisected at O.
In ΔODP and ΔOBQ,
∠BOQ = ∠POD ...[Since, vertically opposite angles]
∠OBQ = ∠ODP ...[Alternate interior angles]
And OB = OD ...[Given]
∴ ΔODP ≅ ΔOBQ ...[By ASA congruence rule]
∴ OP = OQ ...[By CPCT rule]
So, PQ is bisected at O.
Hence proved.
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