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Question
In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: QAP is a straight line.
Solution
Since BE and CF are medians,
F is the mid-point of AB and E is the mid-point of AC.
Now, the line joining the mid-point of any two sides is parallel and half of the third side, we have
In ΔACQ,
EF || AQ and EF = `(1)/(2)"AQ"` ....(i)
In ΔABP,
EF || AP and EF = `(1)/(2)"AP"` ....(ii)
From (i) and (ii), we get AP || AQ (both are parallel to EF)
As AP andAQ are parallel and have a common point A, this is possible only if QAP is a straight line.
Hence proved.
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