In $△ A B C, O B, O C$ are angular bisectors and $M N ∥ B C$ , if $A B = 12$ , $B C = 24$ and $A C = 18$ , then the perimeter of $△ A M N$ is

30

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33

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36

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39

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Solution

The correct option is A $30$
$△ A M N \Rightarrow ?$
$∠ M B O = ∠ C B O = ∠ M O B = \frac{1}{2} ∠ M B C$
$A M + M N + N A = A M + M O + N O + N A$
$= A M + M B + N C + N A$
$= A B + A C = 30$

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¯ ¯¯¯¯¯¯ ¯ B C

¯ ¯¯¯¯¯¯ ¯ A C

¯ ¯¯¯¯¯¯¯¯ ¯ A M

¯ ¯¯¯¯¯¯¯¯¯ ¯ M N + ¯ ¯¯¯¯¯¯¯ ¯ N P

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)

. 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)

In fig., if OC = 9 cm and OB = 15 cm, then BC + BD is equal to:

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

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