ABCD is a Parallelogram E and F are the mid-Points of BC and AD respectively. Show that the
segments BF and DE trisect the diagonal AC.
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In a parallelogram ABCD, E and F are the mid-points segments AF and EC trisect the diagonal BD.
Answer: ABCD is a parallelogram.∴AB || CDAnd hence, AE || FCAgain, AB = CD (Opposite sides of parallelogram ABCD)AB =CDAE = FC (E and F are mid-points of side AB and CD)In quadrilateral AECF, one pair of opposite sides (AE parallelogram.⇒ AF || EC (Opposite sides of a parallelogram)In ΔDQC, F is the mid-point of side DC and FP || CQ said that P is the mid-point of DQ.⇒ DP = PQ ... (1)Similarly, in ΔAPB, E is the mid-point of side AB and it can be said that Q is the mid-point of PB.
⇒ PQ = QB ... (2)From equations (1) and (2),DP = PQ = BQHence, the line segments AF and EC trisect the diagonal.
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