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Question

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). The line segments AF and EC _____ the diagonal BD.

  1. Trisect
  2. Bisect
  3. Simply intersect at two points
  4. None of these


Solution

The correct option is A Trisect
Given: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD respectively.
To Prove: Line segments AF and EC trisect the diagonal BD.
Proof: Since, ABCD is a parallelogram,
AB || DC
And, AB = DC (Opposite sides of a parallelogram)
AE || FC and 12 AB = 12 DC
AE || FC and AE = FC
AECF is a parallelogram
AF || EC
EQ || AP and FP || CQ
In Δ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid - point of BP.
(By converse of mid-point theorem)
BQ = OQ ........(i)
Again, in  Δ DQC,F is the mid-point of DC and FP || CQ, so P is the mid - point of DQ.
(By converse of mid-point theorem)
QP = DP  ….(ii)
From Equations (i) and (ii) , we get
BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD.

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