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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 enter image description here

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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 $HA=HO=3\cdot GO$.

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  • $\begingroup$ Thanks Dan, but I don't understood : HAC = 90 - C = HBC . Can you explain better? If possible, draw your solution, $\endgroup$ Nov 9, 2019 at 13:41
  • $\begingroup$ I will use the inserted picture from the OP. Here, the angle $\widehat{HBC}=\widehat{H'BC}$ complements the angle in $C$ in the triangle $\Delta H'BC$, since $BH'$ is the height in $\Delta ABC$. So $\widehat {H'BC}+\underbrace{\widehat {H'CB}}_{=\hat C}=90^\circ$. Same applies also for the situation of $HA$ instead of $HB$ (although the full height is not drawn). $\endgroup$
    – dan_fulea
    Nov 9, 2019 at 23:13

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