geometry
In a triangle ABC, H, G, and O are orthocenter, centroid, and circumcenter of the triangle. If the Euler's line is parallel AC and m <(HBC) = 2m <(OCA), calculate GO if AH = a (answer: a/3)
I tried to draw the triangle and relate the properties but couldn't find a solution.
We know that GH = 2GO and BG = 2GP
Triangle BHG ~ POG
This is trivial.
Since $X$ divide $HO$ in ratio $2:1$ point $X$ is actually gravity center of $ABC$, so $X$ lies on $AD$ and divide it in ratio $2:1$.
But $AD$ is median for triangle $AOO'$ so $X$ is also gravity center for $AOO'$.
Let $|BC|=a$, $|AC|=b$, $|AB|=c$, $|GE|=|DE|$.
The distances to the tangent point $G$ of the incircle are
\begin{align} |BG|&=\tfrac12(a+c-b) \tag{1}\label{1} ,\\ |CG|&=\tfrac12(a+b-c) \tag{2}\label{2} , \end{align}
and by the power of the point $G$ w.r.t the circumcircle,
\begin{align} |BG|\cdot|CG|&=|DG|\cdot|EG|=|DG|^2=100 \tag{3}\label{3} ,\\ |BG|\cdot|CG|&=\tfrac14(a+c-b)(a+b-c) =\tfrac14(a^2-(c-b)^2) =\tfrac14(b^2+c^2-(c-b)^2) =\tfrac12\,bc \tag{4}\label{4} , \end{align}
hence, the area of $\triangle ABC$ is $100$.
Best Answer
We have the equality of (measures of) angles: $$ \widehat{HAC} =90^\circ -\hat C=\widehat{HBC}=2\cdot\widehat{OCA} = 2\cdot\widehat{OAC}\ . $$ This implies: $$ \widehat{HAO} = \widehat{OAC} = \widehat{AOH}\ . $$ The triangle $\Delta HAO$ is thus isosceles in $H$, so $HA=HO=3\cdot GO$.