Prove that for any triangle, H (orthocenter), G (centroid), and O (circumcenter) are collinear, and prove that

HG = 2GO.

First, explore using the following file.

Drag the vertices of given triangle and observe the change of length HG and GO.

Now, I will prove the property, HG = 2 GO.

To prove this, we need to show the following Lemma.

<Lemma>

For given triangle ABC, let perpendicular foot of O (circumcenter) and H (orthocenter) be M and D, respectively. Then,

   AH  = 2 OM

<Proof>

First, draw circumcircle of the triangle ABC.

Let the intersection of extension of BC and the circumcircle be R, the intersection of extension of BH and the extension of AC be F, and the perpendicular foot of C be E.

Since BR is a diameter of the circle,

.

Since AR and CE  are perpendicular to AB or extension of AB,

.......(1) .

Since and ,

  ......(2)

By (1) and (2), opposite sides of the quadrilateral ARCH are parallel, so ARCH is a parallelogram. Thus,

.... (3).

Since triangle BOM and BRC are similar with the ratio 1:2,

... (4)

By (3) and (4) ,

AH = 2 OM

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Now, let's prove HG=2GO.

Let perpendicular foot of O (circumcenter) and H (orthocenter) be M and D, respectively.

Consider triangle OMG and HAG.

Since OM // AH (by previous lemma),

....(1)

Since G is the centroid of the triangle ABC,

MG : AG = 1 : 2     .....(2)

By Lemma,

OM : AH = 1 : 2     ....(3)

By (1), (2), and (3), triangle OMG and HAG are similar triangle with ratio 1 : 2.

Therefore,

OG : HG = 1 : 2

that is, HG = 2 OG.

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