Question

In figure, $O$ is an interior point of $\Delta ABC$. Show that: $AB+BC+CA<2(OA+OB+OC)$.

Answer 1

In triangle $ABC$, $O$ is a point interior of $\Delta ABC$.
As we know that “The sum of any two sides of a triangle is greater than the third side”.
$OA+OB>AB$ …(i)
$OA+OC>AC$ …(ii)
and $OB+OC>BC$ …(iii)
Now, adding (i), (ii) and (iii), we get
$2(OA+OB+OC)>AB+BC+CA$
or $AB+BC+CA<2(OA+OB+OC)$
Hence proved.