Question

In the adjoining figure,ABCD is a trapezium in which AB||DC and its diagonals AC and BD intersect at O.

Prove that $ar\left(△AOD\right)=ar\left(△BOC\right)$.

Answer 1

$△CDA)$ and $△CBD)$​ lies on the same base and between the same parallel lines.
So, $ar\left(△CDA\right)=ar\left(△CDB\right)$ ...(i)
Subtracting ar(​∆OCD) from both sides of equation (i), we get:
ar(​∆CDA) − ar(​∆OCD) = ar(​​∆CDB) - ar (​​∆OCD)
⇒ ar(​​∆AOD) = ar(​​∆BOC)