Question

In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.​

Answer 1

● ANSWER

ABCD is ∥gm

AB ∥ CD

AE ∥ FC

⇒AB = CD

1 / 2 AB = 1 / 2 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 .

HOPE IT HELPS YOU !!

Answer 2

Answer:

hope it's help full for u