Prove that centroid, orthocenter and circumcenter are in the ratio 2:1.!!
my attempt..
I could prove they are in the ratio 2:1.Assuming that they are collinear.But couldn't prove that they are collinear.I need help?
[Math] Prove that centroid, orthocenter and circumcenter are collinear
geometry
Related Solutions
- in $\bigtriangleup$ABC AB=AC
- WE take AD$\perp$BC.clearly BD=DC & $\angle$BAD=$\angle$DAC
- so clearly the circumcentre;orthocentre;incentre and centroid - all of them lie on the line AD
- SO that is prooved
since $$\angle HAO=|B-C|.OA=R,AH=2R\cos{A}$$ use cosin theorem we have \begin{align*} OH^2&=AH^2+AO^2-2AH\cdot AO\cos{|B-C|} =4R^2\cos^2{A}+R^2-4R^2\cos{A}\cos{(B-C)}\\ &=5R^2-4R^2\sin^2{A}+4R^2\cos{(B+C)}\cos{(B-C)}\\ &=9R^2-4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C})\\ &=9R^2-(a^2+b^2+c^2) \end{align*}
Best Answer
The circumcenter $O$ of $ABC$ is just the orthocenter of the medial triangle $A'B'C'$, which, obviously, shares its centroid $G$ with $ABC$. Since $\frac{AG}{GA'}=\frac{BG}{GB'}=\frac{CG}{GC'}=2$, $G,O,H$ are collinear and $\frac{HG}{GO}=2$ holds.