Show that the diagonals of a parallelogram divide it into four triangles of equal area. GIVEN : A parallelogram `A B C D` . The diagonals `A C` and `B D` intersect at `Odot` TO PROVE : `a r( O A B)=a r( O B C)=a r( O C D)=a r( A O D)`

Explanation

Given: A parallelogram `ABCD`. The diagonals `AC` and `BD` intersect at `O`
To prove: `ar(/_\AOB)=ar(/_\BOC)=ar(/_\COD)=ar(/_\AOD)`
Proof: Let `ABCD` be a parallelogram with diagonals `AC` and `BD` intersecting at `O`.
Since the diagonals of a parallelogram bisect each other at the point of intersection.
`therefore AO=OC` and `BO=OD`
We know that the median of a triangle divides it into two equal parts.
Now,
In `△ABC`,
`=>BO` is median.
`ar(△AOB)=ar(△BOC)`.....(1)
In `△BCD`,
`=>CO` is median.
`ar(/_\BOC)=ar(/_\COD)`.....(2)
In `/_\ACD`,
`=>DO` is median.
`ar(/_\AOD)=ar(/_\COD)`.....(3)
From equation (1), (2) & (3), we get
`ar(/_\AOB)=ar(/_\BOC)=ar(/_\COD)=ar(/_\AOD)`
Hence Proved.