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Question

# ABCD is a parallelogram in which P is the midpoint of DC and Q is point on AC such that CQ=1/4 AC . If produced PQ meets at BC at R then prove that R is the midpoint of BC.

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Solution

## Given: ABCD is a parallelogram. P is the mid point of CD. Q is a point on AC such that CQ=(1/14)AC PQ produced meet BC in R. To prove : R is the mid point of BC Construction : join BD in O.Let BD intersect AC in O. Prove : O is the mid point of AC. {diagnols of parallelogram bisect each other } Therefore OC = (1/2) AC => OQ = OC-CQ = (1/2)AC - (1/4)AC = (1/4)AC. => OQ = CQ therefore Q is the mid point of OC. In triangle OCD, P is the mid point of CD and Q is the mid point of OC, therefore PQ is parallel to OD (Mid point theorem) => PR is parallel to BD In traingle BCD, P is the midpoint of CD and PR is parallel to BD, therefore R is the mid point of BC (Converse of mid point theorem)

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