Hello. This post is about a pretty fascinating property of triangles. Let’s jump right into it.

As a pre-requisite, you must know the following three special centers of a triangle:

Orthocenter (H): The point of concurrency of the three altitudes.

Centroid (G): The point of concurrency of the three medians.

Circumcenter (O): The point of concurrency of the sides’ perpendicular bisectors.

The vertices can in each applet above can be dragged.

Now, let’s observe them together, in the same triangle. Press the play button in the applet below to start the show. Once again, you can drag the vertices to play around.

What just happened? Seems like the points H, G, O lie on a straight line!

Black magic? Damn right! Much of geometry is black magic.

The three points, i.e. the orthocenter, the centroid, and the circumcenter, are always collinear.

It doesn’t end here. The centroid divides the line segment joining the orthocenter and the circumcenter in the ratio 2 : 1. That is, HG : OG = 2 : 1. Observe the same in the applet below.

And the line joining them is called the Euler line.

Sadly, the incenter (i.e., the point of concurrency of the angle bisectors) isn’t a part of this elite club of collinear points. Maybe it is the rebel, the dark knight, the one. Anyways, to the proof.

Proof

The proof involves a few pre-requisites too:

Congruence of Triangles

The corresponding parts of two congruent triangles are equal.

Similarity of Triangles

The corresponding angles of two similar triangles are equal, and the corresponding sides are proportional.

The Inscribed Angle Theorem

In a circle, the angle subtended by a chord at its center is twice the angle subtended by the chord at the circumference.

A Property of the Centroid

The centroid of a triangle divides each median in the ratio 2 : 1. (Hmm.. 2 : 1 🤔)

Trigonometric Ratios

In ΔABC, where B = 90°, sinA = BC/AC, cosA = AB/AC, and tanA = BC/AB.

The Sine Rule

In a triangle ABC, BC/sinA = CA/sinB = AB/sinC = 2R, where R is the triangle’s circumradius.

With these tools in hand, let’s proceed with the proof. I won’t reveal the whole thing – you’ll have to figure out a few steps on your own.

Have a look at the figure below, which shows the altitude CE, the perpendicular bisector of AB, and the median CD.

Also marked are H and O (i.e. the orthocenter and circumcenter) on the altitude and perpendicular bisector respectively.

Now, CH = CF.sec(90 – B), CF = AC.cosC, and AC = 2R.sinB. In terms of R and C, what does CH simplify to?

Since O is the circumcenter, ∠AOB = 2C. Since ΔAOD and ΔBOD are congruent, ∠DOB = C. Can you find OD in terms of R and C?

Now look at the triangles CHG and DOG 🐕.

Are they similar? ∠HGC and ∠DGO are equal, as they’re vertically opposite. What about ∠HCG and ∠ODG?

(Hint: CE is the altitude and OD is the perpendicular bisector. Think parallel lines and transversals.)

If they’re similar, it means that HG : OG = CH : OD = ? : ?

In case that ratio comes out to be 2 : 1, then congratulations – half of the proof is done.

What would make G the centroid?

The similarity leads to the fact that CG : GD = 2 : 1 too. This means that G is a point on the median, that divides it in the ratio 2 : 1, something that tells us that it must be the centroid!

That’s all for this one. Take some time to figure out the calculations. And of course, drop me a message here if you need some help figuring out.

Further Exploration

By the way, some more special points also lie on this black-magical Euler line – the Exeter point, the nine-point center, and the Schiffler point (apart from many others). Why don’t you find more about these points and why they should lie on the line?

See you in some other post, some other time!