# If S is the circumcenter,G be the centroid,O be the orthocenter of triangle ABC,then SA+SB+SC is equal to

Arun
25750 Points
6 years ago
Dear Student,..In which terms you want the answer?Since from here I can see that SA = SB = SC = R (Circumradius )and is the distance between S and any vertex, which can be easily calculated if verteces or any other info is given.
17 Points
6 years ago
Here,
S=Circumcentre
O=Orthocentre
Consider a triangle ABC. Let D be the perpendicullar bisector of BC drawn from vertex A,similarly Q be the perpendicular bisector of AC drawn from vertex B and thye intersecting point of these bisectors be O.Now S be the intersecting point of all the three medians drawn from all vetex.
Let,$\vec{SA}=\vec{a},\vec{SB}=\vec{b},\vec{SC}=\vec{c},\vec{SO}=\vec{s}$
Now,$\vec{SP} is \parallel to \vec{AO}$
$\Rightarrow$$\vec{SP}=(\vec{SB}+\vec{SC})/2$
$=(\vec{b}+\vec{c})/2$
$\Rightarrow \vec{SP}=\lambda (\vec{AO})$
$\Rightarrow (\vec{b}+\vec{c})/2=\lambda (\vec{s}-\vec{a})$
$\Rightarrow (\vec{b}+\vec{c})/2\lambda +\vec{a}=\vec{s}..........$(1)
Now,$\vec{SQ} is \parallel to \vec{BO}$
$\Rightarrow \vec{SQ}=(\vec{SC}+\vec{SA})/2 \Rightarrow \vec{SQ}=(\vec{a}+\vec{c})/2$
$\Rightarrow \vec{SQ}=\mu \vec{BO}$
$\Rightarrow (\vec{a}+\vec{c})/2=\mu (\vec{s}-\vec{b})$
$\Rightarrow (\vec{a}+\vec{c})/2\mu +\vec{b}=\vec{s}........$(2)
Equate (1) and (2)
$\Rightarrow \vec{a}+(1/2\lambda )\vec{b}+(1/2\lambda )\vec{c}=(1/2\mu )\vec{a}+\vec{b}+(1/2\mu)\vec{c}$
Equating coefficients,we get
$\Rightarrow 1/2\lambda =1$
$\Rightarrow \lambda =1/2$
From (1),
$\Rightarrow \vec{a}+ (\vec{b}+\vec{c})/2\frac{1}{2}=\vec{s}$
$\therefore \vec{a}+\vec{b}+\vec{c}=\vec{s}$
$\therefore \vec{SA}+\vec{SB}+\vec{SC}=\vec{SO}$