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Answer :
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.