$O$ is any point in the interior of $\triangle ABC$. Prove that
$OA + OB + OC > \frac{1}{2}(AB + BC + CA)$
Given:
$O$ is any point in the interior of $\triangle ABC$.
To do:
We have to prove that $OA + OB + OC > \frac{1}{2}(AB + BC + CA)$.
Solution:
Extend $BO$ to meet $AC$ at $D$.
From the figure,
In triangles $OAB, OBC$ and $OCA$,
$OA + OB > AB$........(i)
$OB + OC > BC$.........(ii)
$OC + OA > CA$.........(iii)
Adding equations (i), (ii) and (iii), we get,
$2(OA + OB + OC) > AB + BC + CA$
$OA + OB + OC > \frac{1}{2}(AB + BC + CA)$
Hence proved.
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