Question

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (∆AOD) = ar (∆BOC).

Answer 1

Since ∆ACD and ∆BDC are on the same base CD and between the same parallels AB and CD.

Therefore, ar (∆ACD) = ar (∆BDC)

Subtracting ar (∆ACD) from both side, we get

ar (∆ACD) – ar (∆COD) = ar (∆BDC) – ar (∆COD)

⇒ ar (∆AOD) = ar (∆BOC)