In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.


ABCD is a parallelogram.

AB || CD


And


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 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


Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ


DP = PQ (1)


AB = CD


Similarly,


In ΔAPB,


E is the mid-point of side AB and EQ || AP


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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