Solution:
In ΔABC, AO is the median. (CD is bisected by AB at O)
∴ar(AOC) = ar(AOD) — (i)
also,
ΔBCD, BO is the median. (CD is bisected by AB at O)
∴ar(BOC) = ar(BOD) — (ii)
Adding (i) and (ii),
We get,
ar(AOC)+ar(BOC) = ar(AOD)+ar(BOD)
⇒ar(ABC) = ar(ABD)
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