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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: 2EF = BD.
Solution
In ΔBCD, E and F are the mid-points of DC and BC respectively.
Also EF || BD
Therefore, EF = `(1)/(2)"BD"`
⇒ 2EF = BD.
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