Question

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

Answer 1

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