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The Brainliest Answer!
Consider a ║gm ABCD.
as we know that the diagonals of a ║gm divide it into two triangles of equal areas, so
ar. (ADC) = ar. (BDC)
now, in tri. ADC, OD is the median so,
ar. ( AOD) = ar. (AOB )         ....1
similarly, ar. (COD) = ar. (BOC)       ....2
NOW, ar. (ADC) = ar. (BDC)
⇒ 1/2 ar.(ADC) = 1/2 ar. (BDC)
from 1 and 2,
ar. (AOD) = ar. (AOB) = ar. (COD) = ar. (BOC)
hence, proved.

2 3 2
plz mark as the brainliest....: )
ABCD is a parallelogram in which the diagonals bisect each other

OD is the median 
∴ar DOA = ar DOC ________ (1)

similarly in ΔADB ,OA i s the median 

∴ar DOA = ar BOA________________(2)

similarly we can prove
ar AOB = ar BOC_______________(3)
 ar BOC = ar DOC _______________(4)

from 1 , 2 , 3 and 4

we get

ar (DOC)= ar(DOA)=ar(AOB)= ar(BOC)


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