Question

O is the point in the exterior of $△ABC$.Use the symbols$<,>\text{or}=\text{to}$make the following statements true.

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

Answer 1

Construction of figure;

$\mathrm{O}$ is the point in the exterior of $△\mathrm{ABC}$.

Join $\mathrm{OA},\mathrm{OB}\mathrm{and}\mathrm{OC}$.

Here, we can see $\mathrm{AOB}$ is a triangle.

We know that Triangle inequality theorem states that , Sum of the length of any two sides of a triangle is greater than the third side.

$$\begin{gathered} \mathrm{OA}+\mathrm{OB}>\mathrm{AB} \\ \mathrm{OB}+\mathrm{OC}>\mathrm{BC} \\ \mathrm{OA}+\mathrm{OC}>\mathrm{AC} \end{gathered}$$

Adding all the above equations, we get;

$$\begin{gathered} \mathrm{OA}+\mathrm{OB}+\mathrm{OB}+\mathrm{OC}+\mathrm{OC}+\mathrm{OA}>\mathrm{AB}+\mathrm{BC}+\mathrm{CA} \\ 2(\mathrm{OA}+\mathrm{OB}+\mathrm{OC})>\mathrm{AB}+\mathrm{BC}+\mathrm{CA} \end{gathered}$$

Hence proved.