Question

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Answer 1

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Answer 2

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Given:-ABCD is a parallelogram and E and F are the mid point of sides AB and CD respectively.

=) ∴AB || CD =) Also , AE || FC

To prove:-AF and EC trisect diagonal BD

Proof:-
AB = CD (Opposite sides of parallelogram ABCD)

and AE = FC (E and F are mid-points of side AB and CD)

Now, in quadrilateral AECF , AEIICF and AE=CF

=)AECF is a parallelogram.

⇒ AF || EC (Opposite sides of a parallelogram)

In ΔDQC, F is the mid-point of side DC and FP || CQ

So, by using the converse of mid-point theorem,
=) 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).
So, by using the converse of mid-point theorem

=) Q is the mid-point of PB.

⇒ PQ = QB -------2)

From equations (1) and (2) we get:-

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.


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