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:
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....;^)
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