ABCD is a parallelogram in which E and F are the midpoints of the sides AN and CD respectively. Prove that line segments CE and AF trisects the diagonal BD . PROOF :-
Answers
Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.
To Prove: Line segments AF and EC trisect the diagonal BD.
According to the converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle that is parallel to another side bisects the third side.
ABCD is a parallelogram.
AB || CD
Hence, AE || FC
Again, AB = CD (Opposite sides of parallelogram ABCD)
⇒ 1/2 AB = 1/2 CD
⇒ AE = FC (E and F are mid-points of side AB and CD)
In quadrilateral AECF, one pair of opposite sides (AE and CF) 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 (as AF || EC).
Therefore, by using the converse of the mid-point theorem, it can be said that P is the mid-point of DQ.
∴ DP = PQ --------------- (1)
Similarly, in ΔAPB, we know E is the mid-point of side AB and thus, EQ || AP (as AF || EC).
Therefore, by using the converse of the 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.
Answer:
Step-by-step explanation: ABCD is //gm
AB is parallel to CD
AE is parallel to FC
=AB is parallel to CD
1/2AB=1/2CD
AE=EC
in triangle DCQ
F is mid point of DC
FP is parallel to CQ
By converse of mid point theorem P is mid point of DQ
=DP=PQ (1)
Therefor AF and EC bisect BD
in triangle APB
E is mid point of AB
EQ is parallel to AP
by convers of MPT (mid point theorem)
Q is mid point of PB
=PQ=QB (2)
By (1) and (2)
= PQ=QB=DP
AF and EC bisect BD..