Question

Only for Moderators and maths genius.


In a parallelogram ABCD ,E and F are the mid-points of sides AB and CD respectively. Show that the Line segment AF and EC bisect the diagonal BD.


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Answer 1

Answer:

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..

Answer 2

Answer:

Step-by-step explanation:Answer:  

LCM of 3,6,9=18

Now let us change each of the given fraction into an equivalent fraction having 18 as the denominator.

(2/3)x(6/6)=(12/18)

and (5/6)x(3/3)=(15/18)

and (1/9)x(2/2)=(2/18)

Now,

=(12/18)+(15/18)−(2/18)

=(12+15−2)/18

=(27−2)/18

=(25/18)

=[1(7/18)]