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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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GIVEN A quad. ABCD in which diagonals AC and BD are such that

`ar(triangleABD)=ar(triangleCDB)and ar(triangleABC)=ar(triangleACD).`
TO PROVE ABCD is a parallelogram.
PROOF `ar("quad. ABCD")=ar(triangleABC)+ar(triangleACD)`
`=2ar(triangleABC)" " [therefore (triangleACD)=ar(triangleABC)]`.
`therefore ar(triangleABC)=(1)/(2)ar("quad. ABCD")." "...(i)`
Again, `ar("quad. ABCD") = ar(triangleABD)+ar(triangleCDB)=2ar(triangleABD)`
`[therefore ar(triangleCDB)=ar(triangleABD)]`.
`therefore ar(triangleABD)=(1)/(2)ar("quad. ABCD")." "...(ii)`
From (i) and (ii), we get
`ar(triangleABC)=ar(triangleABD)`. Also, `triangle ABC and triangleABD` have the same base AB.
`therefore` they lie between the same parallels AB and DC [ Theorem 9 ]
i.e., AB||DC.
Similarly, AD||BC.
Hence, ABCD is a ||gm.
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