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# 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.