Question

parallelogram ABCD E and F are the midpoints of sides ab and CD respectively show that the line segment AF and EC trisect the diagonal BD

Answer 1

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively.

To Prove: Line segments AF and EC trisect the diagonal BD.

Proof:

AB = DC | Opposite sides of ||gm ABCD

| Halves of equals are equal

AE = CF ....(2)

In view of (1) and (2),

AECF is a parallelogram

| A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length

∴ EC || AF ...(3)

| Opposite sides of || gm AECF

In ∆DQC,

∵ F is the mid-point of DC

and FP || CQ | ∵ EC || AF

∴ P is the mid-point of DQ

| By converse of mid-point theorem

⇒ DP = PQ ...(4)

Similarly, in ∆BAP,

BQ = PQ ...(5)

From (4) and (5), we obtain

DP = PQ = BQ

⇒ Line segments AF and EC trisect the diagonal BD.

Answer 2

Given :-

ABCD is a parallelogram.

E and F are the midpoints of AB and CD

To be proof :

DP = PQ = QB

Construction : Join AF and CE

PROOF :

Since,

ABCD is a parallelogram,

So,

AB║CD and AB = CD

AE║CD and AB = CD

∴ AE║FC

and AE = FC     [ AE = ¹/₂ AB, FC = ¹/₂ CD ]

⇒ AFCE is a parallelogram.

[ a pair of opposite sides are equal and parallel ]

Now,

In ΔDQC,

FP║CQ and F is the midpoint of CD

PQ = DP  ..............(i)

Similarly from ΔAPB, Q is the midpoint of BP

So,

PQ = BQ   ..............(ii)

From (i) and (ii),

PQ = DP = BQ

Hence,

AF and CF trisect the diagonal BD.