Prove that $O A + O B + O C < A B + B C + C A$ if $O$ is the interior point.

Open in App

Solution

Let $O$ be the interior point of $Δ A B C$ .
Also let $B O$ is produced to intersect $A B$ at $D$ .
Now for
$\begin{matrix} Δ A B D A B + A D > B D \Rightarrow A B + A D > B O + O D - - - (1) F o r, Δ C O D C D + O D > O C - - - - - (2) o n a d d i n g (1) a n d (2) w e g e t A B + A C > O B + O C - - - (3) s i m i l a r l y A B + B C > O A + O C - - - (4) a n d A C + B C > O A + O B - - - - (5) a d d i n g (3) (4) a n d (5) w e g e t 2 (A C + B C + C A) > 2 (O A + O B + O C) ∴ O A + O B + O C < A B + B C + C A p r o v e d . \end{matrix}$

Suggest Corrections

Similar questions

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

Prove that

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

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

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

. 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)

Q. If any point

O

in the interior of a triangle ABC. Prove that AB

+

+

>

+

+

OC.

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

$If O is a point within Δ A B C, show that (i) AB + AC > O B + O C (ii) AB + BC + CA > O A + O B + O C (iii) OA + OB + OC > \frac{1}{2} (A B + B C + C A .)$

Join BYJU'S Learning Program

Prove that OA+OB+OC<AB+BC+CA if O is the interior point.

Prove that $O A + O B + O C < A B + B C + C A$ if $O$ is the interior point.