In fig, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar(ABC) = ar(ABD).

In ΔABC, AO is the median. CD is bisected by AB at O. Therefore, ar(AOC) = ar(AOD) — (i)
Also, In ΔBCD, BO is the median. CD is bisected by AB at O. Therefore, ar(BOC) = ar(BOD) — (ii)
Adding (i) and (ii) we get, ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD) ⇒ ar(ABC) = ar(ABD).