Question

$O$ is any point in the interior of $\Delta ABC.$ then
$AB+BC+CA<OA+OB+OC$ that is ?

• A. True
• B. False

Answer 1

The correct option is B False

Let $O$ be a inner point of a $△ABC.$

$BO$ is produced to intersect $AB$ at $D.$

In $△ABD$

$AB+AD>BD$

$\Rightarrow$ $AB+AD>BO+OD$ ----- ( 1 )

In $△COD,$

$CD+OD>OC$ ------ ( 2 )

Adding ( 1 ) and ( 2 ) we get,

$AB+AD+CD+OD>BO+OD+OC$

$\Rightarrow$ $AB+AC+OD>BO+OD+OC$

$\Rightarrow$ $AB+AC>OB+OC$ ------ ( 3 )

Similarly,

$\Rightarrow$ $AB+BC>OA+OC$ ----- ( 4 )

And

$\Rightarrow$ $AC+BC>OA+OB$ ----- ( 5 )

Adding ( 3 ), ( 4 ) and ( 5 ) we get,

$2(AB+BC+CA)>2(OA+OB+OC)$

$\Rightarrow$ $AB+BC+CA>OA+OB+OC$

$\therefore$ Given statement in question is false.