Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that
ar ( AOD ) = ar ( BOC ). Prove that ABCD is a trapezium....!!
I GuARAnTEeEE ThUMBS uP* FoR De CrrCt aNsaZz... ;)
It is given that AOD = BOC . Now, add DOC in both triangles
This will give ACD= BCD.
Now there is a theorem that parallelograms on the same base and between the same parallels have equal area.
Now in this condition , The areas of both triangles are equal and they are also on the same base DC so they should be between the same parallels according to the theorem.
THis way AB is parallel to DC.One pair of opposite sides is parallel .
Hence it is proved that ABCD is a trapezium.
It is given thatArea (ΔAOD) = Area (ΔBOC)Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)Area (ΔADB) = Area (ΔACB)We know that triangles on the same base having areas equal to each other lie between the same parallels.Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels.i.e., AB || CDTherefore, ABCD is a trapezium