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Question
ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is the midpoint of BC.
Solution
We know that the diagonals of a parallelogram bisect each other.
Therefore,
`CS=1/2AC` ................(1)
Also, it is given that` CQ=1/4 AC ` .............(2)
Dividing equation (ii) by (i), we get:
`CQ/CS=((1/4)AC)/((1/2)AC)`
Or, CQ=`1/2`CS
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in ΔCSD
PQ || DS
If PQ || DS, we can say that QR || SB
In Δ CSB, Q is midpoint of CS and QR ‖ SB.
Applying converse of midpoint theorem , we conclude that R is the midpoint of CB.
This completes the proof.
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