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.

Question

Philoid · Accepted Answer

It is given that:

Area (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 || CD

Therefore, ABCD is a trapezium