If each diagonals of a quadrilateral separates it into two triangles of equal area then show that the quadrilateral is a parallelogram.

Explanation

With respect to diagonal `AC`,

`⇒ ar(/_\ABC)=ar(/_\ACD)…(i)`

`⇒ ar(/_\ABC)+ar( /_\ACD)=ar(ABCD)…(ii)`
From `(i)` and `(ii)`,

`⇒2ar(/_\ABC)=ar(ABCD)…(iii)`
With respect to diagonal `BD`,
`⇒ar(/_\ABD)=ar(/_\BCD)…(iv)`

`⇒ar(/_\ABD)+ar(/_\BCD)=ar(ABCD)…(v)`
From `(iv)` and (v),
`⇒2ar(/_\ABD)=ar(ABCD)…(vi)`
From `(iii)` and `(vi)` we get,
`⇒2ar(/_\ABC)=2ar(/_\ABD)`

`⇒ar(/_\ABC)=ar(/_\ABD)`
Since `/_\ABC` and `/_\ABD` are on the same base `AB` and have equal area. Therefore, they must have equal corresponding altitudes.
i.e. Altitude from `C` of `/_\ABC=` Altitude from `D` of `/_\ABD`
`⇒ DC∥AB`
Similarly, we can prove that
`AD∥BC`.
So, opposite sides of quadrilateral `ABCD` are parallel which implies that `ABCD` is a parallelogram.