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Question
A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.
Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
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Solution
For help, we draw the diagonal BD as shown below
The diagonal AC and BD cuts at point X.
We know that the diagonal of a parallelogram intersect equally with each other. Therefore
AX = CX and BX = DX
Given,
CQ = `[1]/[4]`AC
CQ = `[1]/[4]` x 2CX
CQ = `[1]/[2]`CX
Therefore Q is the midpoint of CX.
(i) For triangle CDX PQ || DX or PR || BD
Since for triangle CBX
Q is the midpoint of CX and QR || BX. Therefore R is the midpoint of BC
(ii) For triangle BCD
As P and R are the mid-point of CD and BC, therefore PR = `[1]/[2]` DB
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