Question

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

Answer 1

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.