Question

In a parallelogram ABCD, E and
Fare the mid-points of sides AB
and CD, respectively (see the
figure). Show that the line
segments AF and EC trisect the
diagonal BD.

Answer 1

Step-by-step explanation:

ANSWER

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..