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.

• A. Trisect
• B. Bisect
• C. Simply intersect at two points
• D. None of these

Answer 1

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)
$\Rightarrow$ AE || FC and $\frac{1}{2}$ AB = $\frac{1}{2}$ DC
$\Rightarrow$ AE || FC and AE = FC
$\therefore$ AECF is a parallelogram
$\therefore$ AF || EC
$\Rightarrow$ EQ || AP and FP || CQ
In $\Delta$ 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 $\Delta$ 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.