In the figure,$ABC$ and $ABD$ are two triangles on the same base $AB$. If line segment $CD$ is bisected by $AB$ at $O$,show that $\operatorname{ar}(ABC) = \operatorname{ar}(ABD)$.

(N/A) We have $\Delta ABC$ and $\Delta ABD$ on the same base $AB$.
Since $CD$ is bisected at $O$,we have $CO = DO$.
In $\Delta ACD$,$AO$ is a median because it divides the side $CD$ into two equal parts $(CO = DO)$.
Therefore,$\operatorname{ar}(\Delta AOC) = \operatorname{ar}(\Delta AOD)$ (Since a median of a triangle divides it into two triangles of equal areas) $(1)$.
Similarly,in $\Delta BCD$,$BO$ is a median.
Therefore,$\operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta BOD)$ $(2)$.
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta AOC) + \operatorname{ar}(\Delta BOC) = \operatorname{ar}(\Delta AOD) + \operatorname{ar}(\Delta BOD)$
$\Rightarrow \operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta ABD)$.