Question

In $△ABC$, if ${}^{\prime }{O}^{\prime }$ is the circumcentre and $H$ is the orthocentre, then show that $HA+HB+HC=2HO$.

Answer 1

we know that

$HG=2GO$ where $G$ is centroid of triangle

let a point D between B and C

$OD=(OB+OC)/2$

$OA+OB+OC=OA+2OD$

we know That $G$ divide The Point A and midpoint of

opposite side in ratio 2 :1

$OG=\frac{OA+2OD}{3}$

$OA+OB+OC=30G=20G+OG$

$=HG+OG$

$OA+OB+OC=O$

$\overline{HA}+\overline{HB}+\overline{HC}=\overline{HA}+2\overline{HD}=\overline{AA}+2(\overline{HO}+OD)$

$=\overline{HA}+2\overline{HO}+2\overline{OD}$

$=\overline{HA}+\overline{AH}+2\overline{HO}=2HO$