In a parallelogram abcd e and f are the midpoints of sides ab and cd
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ABCD is a parallelogram.
So, AB || CD
⇒ AE || FC
Also AB = CD (Opposite sides of parallelogram ABCD)
⇒ AE = FC (Since E and F are midpoints of AB and CD)
In quadrilateral AECF, one pair of opposite sides are equal and parallel.
∴ AECF is a parallelogram.
Then,
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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