NCERT Solutions
Class 9 Maths
Quadrilaterals

Ex.8.2 Q.5
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.
ABCD is a parallelogram.
So, AB || CD
And hence, AE || FC
Again, AB = CD
(Opposite sides of parallelogram ABCD)
=> =
=> 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 ∆DQC,
F is the mid-point of side DC and FP || CQ
(As, AF || EC).
Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.
So, DP = PQ ...............(1)
Similarly, in ∆APB,
E is the mid-point of side AB and EQ || AP
(As, AF || EC).
Therefore, by using the converse of mid-point theorem, it can be said that Q is the mid-point of PB.
So, PQ = QB ...............(2)
From equations (1) and (2),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.