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Hii There!!
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.
Hope it helps
Mark as Brainliest
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.
Hope it helps
Mark as Brainliest
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