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

Solution:

Consider ΔACD.

Line-segment CD is bisected by AB at O. Therefore, AO is the median of

$\triangle \mathrm{ACD}$

$\therefore$ Area $(\Delta \mathrm{ACO})=$ Area $(\triangle \mathrm{ADO}) \ldots$(1)

Considering $\triangle B C D, B O$ is the median.

$\therefore$ Area $(\Delta \mathrm{BCO})=$ Area $(\Delta \mathrm{BDO}) \ldots(2)$

Adding equations (1) and (2), we obtain

Area $(\Delta \mathrm{ACO})+$ Area $(\Delta \mathrm{BCO})=$ Area $(\triangle \mathrm{ADO})+$ Area $(\triangle \mathrm{BDO})$

$\Rightarrow$ Area $(\triangle \mathrm{ABC})=$ Area $(\triangle \mathrm{ABD})$