Question

ABCD is a parallelogram . Through a point O on the diagonal AC,EOG,FOH are drawn parallel to AB and BC, Ebeing on AD amd F being on AB. Prove that the parallelograms EOHD and FOGB are equal in area.

Answer 1



Ashutosh Verma answered this
in Math, Class

Answer :

We have a parallelogram ABCD , So AB  =  CD and BC  =  DA

And we know diagonal of parallelogram bisect each other , We assume these diagonal bisect at " P "
So
AP  =  PC   And  BP  =  PD                               ----------- ( 1 )

So we form our diagram , As  :

We know diagonal of parallelogram divide it in four same area part . SO

Area of ∆APB  = Area of ∆APD                                                          ---------------- ( A )

For ∆ POB and ∆ POD
And We know from equation 1

BP  =  PD

And Height of both triangle ( ∆ POB and ∆ POD ) will be same As they have same vertex " P "

So let height of both triangle  =  h

We know Area of triangle  = 12× Base × Height
So,
Area of ∆ POB  = 12× BP × h                                            --------- ( 2 )
And
Area of ∆ POD  = 12× PD × h      
We know BP  =  PD , So
Area of ∆ POD  = 12× BP × h      , So from equation 1 , we get

Area of ∆ POB  = Area of ∆ POD                                                       ----------- ( B )

And

Area of ∆ AOB   = Area of ∆ APB  + Area of ∆ POB 
And
Area of ∆ AOD   = Area of ∆ APD  + Area of ∆ POD 

So from equation A and B , we get

Area of ∆ AOB   = Area of ∆AOD                                                                                     ( Hence proved )