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Question
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: F is the mid-point of BC.
Solution
Join B and D. Suppose AC and BD cut at O. Then,
OC = `(1)/(2)"AC"`
Now,
PC = `(1)/(4)"AC"`
⇒ PC = `(1)/(2)"OC"`
In ΔDCO, E and P are the mid-points of DC and OC respectively.
∴ EP || DO
Also, in ΔCOB, P is the midpoint of OC and PF || DO || BD
Therefore, F is the mid-point of BC, F being EP produced.
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