$O$ is any point in the interior of $\triangle ABC$. Prove that
$AB + BC + CA > OA + OB + OC$
Given:
$O$ is any point in the interior of $\triangle ABC$.
To do:
We have to prove that $AB + BC + CA > OA + OB + OC$.
Solution:
Extend $BO$ to meet $AC$ at $D$.
From the figure,
$AB + AD > BD$ (Sum of any two sides of a triangle is greater than the third side)
$AB + AD > BO + OD$.......…(i)
Similarly,
In $\triangle ODC$,
$OD + DC > OC$.......…(ii)
Adding equations (i) and (ii), we get,
$AB + AD + OD + DC > OB + OD + OC$
$AB + AD + DC > OB + OC$
$AB + AC > OB + OC$.......(iii)
Similarly,
$BC + AB > OA + OC$.........(iv)
$CA + BC > OA + OB$...........(v)
Adding equations (iii), (iv) and (v), we get,
$2(AB+BC+CA)>2(OA+OB+OC)$
$AB+BC+CA>OA+OB+OC$
Hence proved.
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