In the adjoining figure, ABC and ABD are two triangles on the same base AB.

Line segment CD is bisected by AB at O                   (Given)
CO = OD                                .....(1)
In ΔCAO, 
AO is the median.                (From (1))
So, arΔCAO = arΔDAO          .....(2)
Similarly, 
In ΔCBD, 
BO is the median                 (From (1))
So, arΔCBO = arΔDBO          .....(3)
From (2) and (3) we have
arΔCAO + arΔCBO = arΔDBO + arΔDAO
⇒">⇒⇒ar(ΔABC) = ar(ΔABD)