$O$ is any point in the interior of $\triangle ABC$. Prove that
$AB + AC > OB + OC$
Given:
$O$ is any point in the interior of $\triangle ABC$.
To do:
We have to prove that $AB + AC > OB + OC$.
Solution:
Extend $BO$ to meet $AC$ at $D$.
From the figure,
$AB + AD > BD$ (Sum of any two sides of a triangle is greater than the third side)
$AB + AD > BO + OD$.......…(i)
Similarly,
In $\triangle ODC$,
$OD + DC > OC$.......…(ii)
Adding equations (i) and (ii), we get,
$AB + AD + OD + DC > OB + OD + OC$
$AB + AD + DC > OB + OC$
$AB + AC > OB + OC$
Hence proved.
Related Articles
- $O$ is any point in the interior of $\triangle ABC$. Prove that$AB + BC + CA > OA + OB + OC$
- $O$ is any point in the interior of $\triangle ABC$. Prove that$OA + OB + OC > \frac{1}{2}(AB + BC + CA)$
- $ABC$ is an isosceles triangle with $AB=AC$. Bisectors of $\angle B$ and $\angle C$ meet at $O$. Join $AO$. Prove that,$OB=OC$ and $AO$ bisect $\angle A$.
- ABC is an isosceles triangle with AC = BC. If $AB^2 = 2AC^2$. Prove that ABC is a right triangle.
- $ABC$ is a triangle and $D$ is the mid-point of $BC$. The perpendiculars from $D$ to $AB$ and $AC$ are equal. Prove that the triangle is isosceles.
- In a $\triangle ABC$, it is given that $AB = AC$ and the bisectors of $\angle B$ and $\angle C$ intersect at $O$. If $M$ is a point on $BO$ produced, prove that $\angle MOC = \angle ABC$.
- In $\triangle ABC$, side $AB$ is produced to $D$ so that $BD = BC$. If $\angle B = 60^o$ and $\angle A = 70^o$, prove that $AD > AC$.
- $ABC$ is a triangle. $D$ is a point on $AB$ such that $AD = \frac{1}{4}AB$ and $E$ is a point on $AC$ such that $AE = \frac{1}{4}AC$. Prove that $DE =\frac{1}{4}BC$.
- In $\triangle ABC, BD \perp AC$ and $CE \perp AB$. If $BD$ and $CE$ intersect at $O$, prove that $\angle BOC = 180^o-\angle A$.
- In the given figure, $ABC$ is a triangle in which $\angle ABC = 90^o$ and $AD \perp CB$. Prove that $AC^2 = AB^2 + BC^2 - 2BC \times BD$"
- In the given figure, $O$ is a point in the interior of a triangle ABC, $OD \perp BC, OE \perp AC$ and $OF \perp AB$. Show that (i) $OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2$.(ii) $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$."
- $D$ is the mid-point of side $BC$ of $\triangle ABC$ and $E$ is the mid-point of $BD$. If $O$ is the mid-point of $AE$, prove that $ar(\triangle BOE) = \frac{1}{8} ar(\triangle ABC)$.
- $ABC$ is a triangle in which $BE$ and $CF$ are respectively, the perpendiculars to the sides $AC$ and $AB$. If $BE = CF$, prove that $\triangle ABC$ is isosceles.
- In the given figure, $ABC$ is triangle in which $\angle ABC > 90^o$ and $AD \perp CB$ produced. Prove that $AC^2 = AB^2 + BC^2 + 2BC \times BD$"
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies