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

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Let AC and BD are two diagonals of quadrilateral ABCD.
Given that, the diagonal AC divides `square ABCD` into two triangles of equal areas.
`:.` area of `Delta ABC = " area of " DeltaACD`
`rArr` area of `Delta ABC = (1)/(2) xx " Area of " square ABCD`...(1)
Similarly, the diagonal BD of `square ABCD` divides it into two triangles of equal areas.
`:.` area of `Delta ABD = " area of " Delta BCD`
`rArr` area of `Delta ABD = (1)/(2) xx " area of " square ABCD`...(2)
From (1) and (2), we get
area of `Delta ABC = " area of " Delta ABD`
But these are on the same base AB. (by theorem 5)
`:.` Their corresponding height will be same.
`rArr AB and DC` are parallel.
Similarly, we can prove that `AD ||BC`
`:. square ABCD` is a parallelogram
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