Question

If O is the circumcentre and P the orthocentre of $\Delta ABC$ , prove that
i) $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OP}$
ii) $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}=2\overrightarrow{PO}$

Answer 1

(i) O-circumcentre

P-orthocentre

$\overline{OA}=a$,

$\overline{OB}=b$,

$\overline{OC}=c$

$\overline{Ob}=\frac{b+c}{2}$

$\therefore OA+OB+OC=\overline{OA}+2\overline{OD}$

$\Rightarrow \overline{OA}+\overline{AP}\Rightarrow \overline{OP}$

[R circumradius of $\Delta$ABC hence $\overline{AH}=\overline{2OD}$]

(ii) $\overline{PA}+\overline{PB}+\overline{PC}=\overline{2PO}$

$\overline{PA}+2\overline{PD}=\overline{PA}+2(\overline{PO}+\overline{OD})$

$=\overline{PA}+2\overline{PO}+2\overline{OD}$

$=\overline{PA}+2\overline{PO}+\overline{AP}$

$=2\overline{PO}$.