Question

ABCD is a parallelogram.E and F are the mid points of side AB and CD respectively. prove that theline segments AF and CE intersect (divide into three equal parts) the diagonal BD

Answer 1

ABCD is a parallelogram.
∴AB || CD
And hence, AE || FC
Again, AB = CD (Opposite sides of parallelogram ABCD)
AB = 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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Answer 2

Please check the attachment. It'll surely help you.