Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Explanation

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.