O if any point in the interior of a triangle ABC. Prove that AB $+$ AC $>$ OB $+$ OC.

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Solution

Construction=produced BO to meet AC in D
$\begin{matrix} o o f = I n Δ A B D = A B + A D > B D = A B + A D > B O + O D \to (i) I n Δ O C D C D + D O > O C \to (i i) A d d i n g e q u a t i o n (i) & 9 i i) w e g e t A B + A C > O B + O C [A D + D C = A C] \end{matrix}$

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Similar questions

. 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 B C

O A + O B + O C > 1 / 2 (A B + B C + A C)

O

+

+

>

+

+

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

O is any point in the interior of $Δ A B C .$

Prove that

$(i) A B + A C > O B + O C$

$(i i) A B + B C + C A > O A + O B + O C$

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

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