O is any point in the interior of ∆ABC. Then, which of the following is true?
(a) $(O A + O B + O C) > (A B + B C + C A)$
(b) $(O A + O B + O C) > \frac{1}{2} (A B + B C + C A)$
(c) $(O A + O B + O C) < \frac{1}{2} (A B + B C + C A)$
(d) None of these

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Solution

(c) $(O A + O B + O C) > \frac{1}{2} (A B + B C + C A)$

In $△ O A B, △ O B C and △ O C A, we have :$
$O A + O B > A B, O B + O C > B C and O C + O A > A C$
Adding them, we get:

$2 (O A + O B + O C) > (A B + B C + C A) \Rightarrow (O A + O B + O C) > \frac{1}{2} (A B + B C + C A)$

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Similar questions

\frac{1}{2}

. O is any point in the interior of a triangle ABC. Prove that

1. AB + AC > OB + OC

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

3.OA + OB + OC > 1/2(AB + AC + BC)

O A + O B + O C > \frac{1}{2} (A B + B C + C A)

O

Δ

A B + A C = O B + O C

A B + B C + C A > O A + O B + O C

O A + O B + O C > \frac{1}{2} (A B + B C + C A)

If $O$ is a point within triangle $A B C$ , show that,

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