Points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Which of the following is correct?

AC and PQ bisect each other.

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AC = CD

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None of the above

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AC = PQ

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Solution

The correct option is A AC and PQ bisect each other.
In triangles AOP and COQ,
$A P = C Q (Given)$
$∠ 1 = ∠ 2 (Alternate interior angles)$
$∠ 3 = ∠ 4 (Vertically opposite angles) ∴ A O P ≅ C O Q (By AAS congruence criterion)$

$∴ O A = O C and O P = O Q (CPCT)$
$Hence, AC and PQ bisect each other.$

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