Question

If $O$ is a point within $\Delta$ABC then show that :
1) $AB+AC=OB+OC$
2) $AB+BC+CA>OA+OB+OC$
3) $OA+OB+OC>\frac{1}{2}(AB+BC+CA)$.

Answer 1

(i) It is given that $O$ is a point within $△ABC$

Consider $△ABC$

We know that $AB+AC>BC..\left(1\right)$

Consider $△OBC$

We know that $OB+OC>BC..\left(2\right)$

By subtracting both the equations we get

$(AB+AC)(OB+OC)>BC-BC$

So we get

$(AB+AC)(OB+OC)>0$

$AB+AC>OB+OC$

Therefore, it is proved that $AB+AC>OB+OC$.

(ii) We know that $AB+AC>OB+OC$

In the same way, we can write

$AB+BC>OA+OC$ and $AC+BC>OA+OB$

By adding all the equations we get

$AB+AC+AB+BC+AC+BC>OB+OC+OA+OC+OA+OB$

So we get

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

Dividing by $2$ both sides

$AB+BC+AC>OA+OB+OC$

(iii) Consider $△OAB$

We know that $OA+OB>AB..\left(1\right)$

Consider $△OBC$

We know that $OB+OC>BC..\left(2\right)$

Consider $△OCA$

$OC+OA>CA\left(3\right)$

By adding all the equations

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

$AB+BC+AC>OA+OB+OC$

(iii) Consider $△OAB$

We know that $OA+OB>AB..\left(1\right)$

Consider $△OBC$

We know that $OB+OC>BC..\left(2\right)$

Consider $△OCA$

$OC+OA>CA\left(3\right)$

By adding all the equations

$OA+OB+OB+OC+OC+OA>AB+BC+CA$.

$OA+OB+OC>\frac{1}{2}(AB+BC+CA)$.