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

AC and PQ bisect each other

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AC and PQ are equal to each other

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AC = CD

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None of these

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Solution

The correct option is A
AC and PQ bisect each other

In AOP and COQ, we have,

AP = CQ [given]

$∠ 1 = ∠ 2$ [ $∵$ alternate interior $∠ s$ ]

$∠ 3 = ∠ 4$ [vertically opp. $∠ s$ ]

$∴ A O P ≅ C O Q$ [by AAS congruence rule]

$∴$ OA = OC and OP = OQ [CPCT]

Hence, AC and PQ bisect each other.

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

The correct option is A AC and PQ bisect each other In AOP and COQ, we have, AP = CQ [given] ∠1=∠2 [∵ alternate interior ∠s] ∠3=∠4 [vertically opp. ∠s] ∴AOP≅COQ [by AAS congruence rule] ∴ OA = OC and OP = OQ [CPCT] Hence, AC and PQ bisect each other.