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Question
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.
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Solution
ABCD is a parallelogram.
∴ AB || CD
And hence, AE || FC
Again, AB = CD ...(Opposite sides of parallelogram ABCD)
`1/2 AB` = `1/2 CD`
AE = FC ...(E and F are mid-points of side AB and CD)
In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram.
⇒ AF || EC ...(Opposite sides of a parallelogram)
In ΔDQC, F is the mid-point of side DC and FP || CQ (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.
⇒ DP = PQ ...(1)
Similarly, in ΔAPB, E is the mid-point of side AB and EQ || AP (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB.
⇒ PQ = QB ...(2)
From equations (1) and (2),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.
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