We have the diagram as
Given: DP = PC &
CQ = (1/4)AC …(i)
To Prove: CR = RB
Proof: Join B to D
As diagonals of a parallelogram bisect each other at S.
Dividing equation (i) by (ii), we get
⇒ CQ = CS/2
⇒ Q is the midpoint of CS.
According to midpoint theorem in ∆CSD, we have
PQ ∥ DS
Similarly, in ∆CSB, we have
QR ∥ SB
Also, given that CQ = QS
We can conclude that, by the converse of midpoint theorem, CR = RB.
That is, R is the midpoint of CB.
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