Question

If $S$ is the circumcentre, $O$ is the orthocentre of $\Delta ABC$, then $\overline{SA}+\overline{SB}+\overline{SC}$ equal to

• A. $\overline{SO}$
• B. $2\left(\overline{SO}\right)$
• C. $\overline{OS}$
• D. $2\left(\overline{OS}\right)$

Answer 1

Answer: (A)

$\overline{SO}$

Consider a $\Delta ABC$. Let $$D$$ be the perpendicular bisector of BC drawn from vertex A. similarly E be the perpendicular bisector of AC drawn from vertex B and they intersecting point of these bisectors be O. now S be the intersecting point of all the three median drawn from all vertex.

Now, let $\overrightarrow{SA}=\overrightarrow{a},\overrightarrow{SB}=\overrightarrow{b},\overrightarrow{SC}=\overrightarrow{c},\overrightarrow{SO}=\overrightarrow{s}$

Now,

$\overrightarrow{SQ}\parallel \overrightarrow{BO}$

$\Rightarrow \overrightarrow{SQ}=\frac{(\overrightarrow{SC}+\overrightarrow{SA})}{2}$

$\Rightarrow \overrightarrow{SQ}=\frac{(\overrightarrow{a}+\overrightarrow{c})}{2}$

But,

$\Rightarrow \overrightarrow{SQ}=\mu \overrightarrow{BQ}$

$\Rightarrow \frac{(\overrightarrow{a}+\overrightarrow{c})}{2}=\mu (\overrightarrow{s}-\overrightarrow{b})$

$\Rightarrow \frac{(\overrightarrow{a}+\overrightarrow{c})}{2\mu }+b=\overrightarrow{s}........\left(1\right)$

Now,

$\overrightarrow{SP}\parallel \overrightarrow{AO}$

$\Rightarrow \overrightarrow{SP}=\frac{(\overrightarrow{SB}+\overrightarrow{SC})}{2}$

$\Rightarrow \overrightarrow{SP}=\frac{(\overrightarrow{b}+\overrightarrow{c})}{2}$

$\Rightarrow \overrightarrow{SP}=\lambda \left(\overrightarrow{AO}\right)$

$\Rightarrow \frac{(\overrightarrow{b}+\overrightarrow{c})}{2}=\lambda (\overrightarrow{s}-\overrightarrow{a})$

$\Rightarrow \frac{(\overrightarrow{b}+\overrightarrow{c})}{2\lambda }+\overrightarrow{a}=\overrightarrow{s}.......\left(2\right)$

By equation (1) and (2) to, we get,

$\Rightarrow \frac{(\overrightarrow{a}+\overrightarrow{c})}{2\mu }+\overrightarrow{b}=$$\frac{(\overrightarrow{b}+\overrightarrow{c})}{2\lambda }+\overrightarrow{a}$

Equating coefficients and we get,

$\frac{1}{2\lambda }=1$

$\lambda =\frac{1}{2}$

By equation (2) to,

$\overrightarrow{a}+(\overrightarrow{b}+\overrightarrow{c})2\times \frac{1}{2}=\overrightarrow{s}$

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

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

Hence, this is the answer.

option$\left(A\right)$ is correct.