Question

In a 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.



Please solve with proof....​

Answer 1

Step-by-step explanation:

ABCD is ∥gm

AB∥CD

AE∥FC

⇒AB=CD

2

1

AB=

2

1

CD

AE=EC

AECF is ∥gm

In △DQC

F is mid point of DC

FP∥CQ

By converse of mid point theorem P is mid point of DQ

⇒DP=PQ (1)

∴AF and EC bisect BD

In △APB

E is mid point of AB

EQ∥AP

By converse of MPT ( mid point theorem )

Q is mid point of PB

⇒PQ=QB (2)

By (1) and (2)

⇒PQ=QB=DP

AF and EC bisect BD..

solution

Answer 2

⛦Given :-

• ABCD is a parallelogram.
• E and F are the midpoints of AB and CD

⛦To prove :-

• 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 EC trisect the diagonal BD.