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(i) ABCD is a parallelogram.
(ii) Diagonals AC and BD bisect each other at O.
(iii) P is the mid-point of DC.
(iv) Q is a point on AC such that CQ = AC
(v) PQ produced meets BC at R.
To Prove:
(i) PQ = DO
(ii) QR = OB
(iii) PQ = QR
(iv) R is the mid-point of BC
Proof: ABCD is a parallelogram.
Therefore, AC and BD bisect each other.
Therefore, OC = OA = AC
But CQ = AC
Therefore, CQ = . AC
= OC
Hence, Q is the mid-point of CO.
In CDO, P is the mid-point of CD and Q is the mid-point of CO. [proved]
Therefore, PQ = DO and PQ || DO [mid-point theorem]
Therefore, part (i) is proved.
Also, as PQ || DO,
QR ||OB.
In COB, Q is the mid-point of CO and QR || OB.
Therefore, R is mid-point of BC. [converse of mid-point theorem]
Therefore, part (iv) proved.
Also, QR = OB, part (ii) proved.
As PQ = DO
QR = OB and DO = OB.
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