O is any point in the interior of . Prove that i. AB+AC>OB+OC ii.… - Vidyadip

Question

$O$ is any point in the interior of $\triangle\text{ABC}.$ Prove that
$i. \text{AB}+\text{AC}>\text{OB}+\text{OC}$
$ii. \text{AB}+\text{BC}+\text{CA}>\text{OA}+\text{OB}+\text{OC}$
$iii. \text{OA}+\text{OB}=\text{OC}>\frac{1}{2}(\text{AB}+\text{BC}+\text{CA)}$

✓

Answer

It is given that, $O$ is any point in the interior of $\triangle\text{ABC}$

We have to prove that.
$i. AB + AC > OB + OC$ produced $BO$ to meet $AC$ at $D.$
In $\triangle\text{ABC}$ we have
$AB + AD > BD$
$\Rightarrow AB + AD > OB + OD ......(1)$
And in $\triangle\text{ODC}$ we have
$OD + CD > OC ......(2)$
Adding $(1) \ (2)$ we get.
$AB + AD + OD + DC > OB + OD + OC$
Hence $AB + AC > OB + OC$ proved.
$ii.$ We have to prove that $AB + BC + CA > OA + OB + OC$
From the first result we have
$BC + BA > OA + OC .......(3)$
And
$CA + CB > OA + OB .......(4)$
Adding above $(4)$ equation.
$2(AB + BC + AC) > 2(OA + OB + OC)$
Hence $AB + BC + CA > OA + OB + OC $proved.
$iii.$ We have to prove that $\text{OA}+\text{OB}+\text{OC}>\frac{1}{2}(\text{AB}+\text{BC}+\text{CA})$
In triangles $OAB, OBC$ and $OCA $ we have
$OA + OB > AB$
$OB + OC > BC$
$OC + OA > AC$
Adding these three results.
$2(OA + OB + OC) > AB + BC + AC$
Hence $\text{OA}+\text{OB}+\text{OC}>\frac{1}{2}(\text{AB}+\text{BC}+\text{CA})$ Proved.

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