Exploration 8:

The Euler Line

John Watson


This exploration is meant to be a guide for exploring properties of the Euler line with the use of Geometer’s Sketch Pad. If you do not have access to GSP, pencil and paper constructions with the aid of a compass and ruler will yield the same results.


Let’s start by constructing a triangle with three arbitrary points X, Y, and Z. (For this exploration, do not make an equilateral triangle)

Now let’s construct the midpoints of each side of the triangle and mark them D, E, F

Let’s draw in the 3 perpendicular bisectors of the triangles 3 sides. The point of concurrency of these lines is called the Circumcenter, (marked C). Notice that it is located at the center of the circle that passes through the three vertices (X, Y, Z).


Let’s draw in the medians. Remember that medians are segments connecting the midpoints with the opposite vertex. The point of concurrency of the medians is called the centroid (marked G).


Next, we’ll construct each of the triangles’ altitudes. Remember that an altitude is the shortest path between a vertex and the line that connects the two other vertices. The Orthocenter (marked H) is the point of concurrency of the lines through each of the triangles vertices that are perpindicular to the lines that extend from the opposite side. The feet of the altitudes are marked L, M, N.

Let’s create new segments that connect the Orthocenter (H) with each of the three vertices (X,Y,Z). Mark the midpoints of these segments with P, Q, R.

Draw a line segment that connects the Orthocenter (H) with the Circumcenter (C). This line is known as the Euler line, named after the Swiss mathematician, Leonhard Euler.

Notice how the Orthocenter (H), Circumcenter (C) and Centroid (G), are all located on the Euler line!

 (Note: If a triangle is equilateral, the Euler line is not defined because the Orthocenter (H), Circumcenter (C), and Centroid (G), will all be located at the same point.)


Another interesting fact is that the distance from the Centroid (G) to the Orthocenter (H) is always twice the distance of the Centroid (G) to the Circumcenter (C). Let’s prove this by using what we know about parallel lines and angles…

Here’s what we already know from our construction…
(i) The line segment from the midpoint D to the circumcenter C is perpendicular to line XZ.

(ii) The segment NY is an altitude so it is also perpendicular to line XZ.

(iii) The Centroid G is 2/3 the distance from vertex Y to the midpoint D on the opposite side. For proof of this phenomenon, check out this exploration by Al Byrnes.

(iv) The Centroid, Circumcenter, and Orthocenter are all on the Euler line.

From (i) and (ii), we know that line DC is parallel to line NY. This means that angle 1 is congruent to angle 4 because they are alternate interior angles.

Similarly, angle 2 and angle 5 are alternate interior angles and therefore congruent.

From (iv) we know that angle 3 and angle 6 are congruent because they are vertical angles.

Since,
Angle 1 is congruent to angle 4
Angle 2 is congruent to angle 5
Angle 3 is congruent to angle 6,
DCG and YHG are similar triangles.

From (iii) we know that the (length of YG) = 2/3 (length of YD).
So, the (length of DG) = (length of YD) – (length of YG) = 1/3 (length of YD).
Therefore, the (length of YG) = 2(length of DG).

Since YG and DG are corresponding sides of similar triangles, we know that the length of any side of triangle YHG is twice the length of the corresponding side of triangle DCG.

Therefore, the (length of GH) = 2(length of GC)!
So, the distance from the Centroid to the Orthocenter (length GH) is equal to twice the distance from the Centroid to the Circumcenter (length GC).

We can confirm this by using the measurement tool in GSP to calculate the (length of GH) and the (length of GC).

When we set up the ratio we can see that

(length of GH) : (length of GC) = 2:1


There are other important points located on the Euler line.  For example, the midpoint between the Circumcenter (C) and the Orthocenter (H) is the center of the 9-point circle, (marked 9pt). Notice how it is located at the center of the blue circle that passes through points: D, E, F, L, M, N, P, Q, R!

Still not convinced? If you are working with GSP try dragging the vertices of the triangle around to see if the above conjectures hold. If you are working with paper and pencil create a new triangle with different dimensions and see if you get the same results.

Click here for more information on the Euler line…


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