Question

Diagonals AC and BD of a trapezium ABCD with AB $\parallel$ DC intersect each other at O.
Prove that ar ($△$AOD) = ar ($△$BOC). [1 MARK]

Answer 1

Concept : 0.5 Mark
Application : 0.5 Mark

It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.

$\therefore$ Area ($△$DAC) = Area ($△$DBC)

$\Rightarrow$ Area ($△$DAC) − Area ($△$DOC) = Area ($△$DBC) − Area ($△$DOC) [ subtracting Area ($△$DOC) on both sides]

$\Rightarrow$ Area ($△$AOD) = Area ($△$BOC)