In Figure, 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 \(\triangle{ACD}\),
Line-segment CD is bisected by AB at O.

So, AO is the median of \(\triangle{ACD}\).

\(\therefore \) Area (\(\triangle{ACO}\)) = Area (\(\triangle{ADO}\)) ...(i)

Similarly,
In \(\triangle{BCD}\), BO is the median.
\(\therefore \) Area (\(\triangle{BCO}\)) = Area (\(\triangle{BDO}\)) ...(ii)

Now, adding equations (i) and (ii), we get,

Area (\(\triangle{ACO}\)) + Area (\(\triangle{BCO}\)) = Area (\(\triangle{ADO}\)) + Area (\(\triangle{BDO}\))
\(\therefore \) Area (\(\triangle{ABC}\)) = Area (\(\triangle{ABD}\))
Hence, proved.