In quadrilateral ABCD, AB = CD [∵ Given] (1/2)AB = (1/2)CD ⇒ AE = CF [∵ E and F are the mid-points of AB and CD respectively] In Quadrilateral AECF, AE = CF [∵ Proved above] AE ∥ CF [∵ Opposite sides of a parallelogram] Hence, AECF is a parallelogram. In ΔDCQ, F is mid-point of DC [∵ Given] and FP ∥Read more

In quadrilateral ABCD,
AB = CD [∵ Given]
(1/2)AB = (1/2)CD
⇒ AE = CF [∵ E and F are the mid-points of AB and CD respectively]
In Quadrilateral AECF,
AE = CF [∵ Proved above]
AE ∥ CF [∵ Opposite sides of a parallelogram]
Hence, AECF is a parallelogram.
In ΔDCQ,
F is mid-point of DC [∵ Given]
and FP ∥ CQ [∵ AECF is a Parallelogram]
Hence, P is mid-point of DQ [ Converse of mid-point theorem]
Hence, DP = PQ …(1)
Similarly,
In ΔABP,
E is mid-point of AB [∵ Given]
and EQ ∥ AP [∵ AECF is a parallelogram]
Hence, Q is mid-point of PB [∵ Converse of mid-Point Theorem]
Hence, PQ = QB …(2)
From (1) and (2), we have
DP = PQ = QB
Hence, line segment AF and EC trisect BD.

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

In quadrilateral ABCD, AB = CD [∵ Given] (1/2)AB = (1/2)CD ⇒ AE = CF [∵ E and F are the mid-points of AB and CD respectively] In Quadrilateral AECF, AE = CF [∵ Proved above] AE ∥ CF [∵ Opposite sides of a parallelogram] Hence, AECF is a parallelogram. In ΔDCQ, F is mid-point of DC [∵ Given] and FP ∥Read more

In quadrilateral ABCD,

See lessAB = CD [∵ Given]

(1/2)AB = (1/2)CD

⇒ AE = CF [∵ E and F are the mid-points of AB and CD respectively]

In Quadrilateral AECF,

AE = CF [∵ Proved above]

AE ∥ CF [∵ Opposite sides of a parallelogram]

Hence, AECF is a parallelogram.

In ΔDCQ,

F is mid-point of DC [∵ Given]

and FP ∥ CQ [∵ AECF is a Parallelogram]

Hence, P is mid-point of DQ [ Converse of mid-point theorem]

Hence, DP = PQ …(1)

Similarly,

In ΔABP,

E is mid-point of AB [∵ Given]

and EQ ∥ AP [∵ AECF is a parallelogram]

Hence, Q is mid-point of PB [∵ Converse of mid-Point Theorem]

Hence, PQ = QB …(2)

From (1) and (2), we have

DP = PQ = QB

Hence, line segment AF and EC trisect BD.