Question

ABCD is a parallelogram in which P is the mid point of CD & CO =1/4 AC. Prove that Ris the
mid point of BC.
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Answer 1

Answer:

Step-by-step explanation:

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 }

∴ OC = (1/2) AC

=> OQ = OC-CQ = (1/2)AC - (1/4)AC = (1/4)AC.

=> OQ = CQ

∴ 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,

∴ R is the mid point of BC (Converse of mid point theorem) (Proved)

If the answer is satisfying,mark it as the brainliest....;^)