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

buddhiraj mishra , 7 Years ago

Grade 12th pass

2 Answers

Arun

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.

Last Activity: 7 Years ago

Satyajit Jadhav

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),

$\Rightarrow \vec{a}+ (\vec{b}+\vec{c})/2\frac{1}{2}=\vec{s}$

$\therefore \vec{a}+\vec{b}+\vec{c}=\vec{s}$

$\therefore \vec{SA}+\vec{SB}+\vec{SC}=\vec{SO}$

Last Activity: 7 Years ago

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