The altitudes and the Euler Line

Existence of the orthocenter can be shown in a variety of ways. Here I would like to present an algebraic proof that invokes the representation of points in the plane as complex numbers. We applied complex numbers to construct similar triangles and concentric polygons, prove Napoleon's theorem, and investigated Morley's complex interpolation theory that led to a spectacular theorem, known as Morley's Miracle.

As the main tool in the proof, I choose the circular coordinates that served Morley in such good stead. In passing, these are also known as Gaussian or conjugate coordinates, the former for historical reasons, the latter in association with the manner in which they are defined.

If X and Y are usual Cartesian coordinates, define x = X + iY and y = X - iY. Conversely, a pair (x,y) (or, actually, either of the conjugate complex numbers x and y) of circular coordinates uniquely determines Cartesian coordinates X and Y. Indeed,

X = (x + y)/2 and Y = (X - y)/2i.

In the Cartesian coordinates a straight line is defined by a linear equation, say

Y = mX + b,

m being the slope, b the y-intercept of the line. (I leave the exceptional case of vertical lines as an exercise.) We can express the same equation in the circular coordinates as

y = Mx + B,

where M = (1 - mi)/(1 + mi) and B = 2bi/(1 + mi). M is the line's clinant, not the slope. The important point is that in the circular coordinates straight lines are again described by linear equations. (Be cautioned, though, that not every linear equation in the circular coordinates describes a line.) For a given point z and its conjugate , the straight line through z with the clinant M is defined by

y = M(x - z) + ,

In the Cartesian coordinates, the line through two points (X1, Y1) and (X2, Y2) has the slope (Y2 - Y1)/(X2 - X1). Similarly (the verification is quite easy), in the circular coordinates, the line through two points (x1, y1) and (x2, y2) has the clinant defined by the ratio (y2 - y1)/(x2 - x1).

The last important fact about the clinants that I should mention is that the clinant of a line perpendicular to a line with a clinant M, is -M. This is because, for m = tan(f),

M = (1 - mi)/(1 + mi)
= (1 - i tan(f))/(1 + i tan(f))
= (cos(f) - i sin(f))/(cos(f) + i sin(f))
= e-2fi

by Euler's and de Moivre's formulas.

But cos(2f + p) = - cos(2f) and sin(2f + p) = - sin(2f), so that e-(2f + p)i = - e-2fi.

We are ready to tackle the problem of concurrency of the altitudes in a triangle. As we'll see, there is a bonus for approaching the problem in the framework of circular coordinates.

Let a triangle be defined by its three vertices x1, x2, and x3. I shall assume existence of the circumcircle: there exists a circle that passes through all three given points. As neither location, nor the "size" of the triangle matter, let's translate and resize it so as to have its circumcenter at the origin and the unit circle xy = 1 as its circumcircle. In other words, let's place the vertices of a triangle on the unit circle. Please note that, with this choice of vertices, the circumcenter is given by x = 0.

The equation of the side x2x3 is

y = M(x - x2) + y2,

where M = (y2 - y3)/(x2 - x3). Since x2 and x3 lie on the unit circle, we can simplify the expression for the clinant M:

M = (y2 - y3)/(x2 - x3)
= (1/x2 - 1/x3)/(x2 - x3)
= (x3 - x2)/(x2 - x3)/(x2x3)
= - 1/(x2x3)
= - y2y3

The equation of the side x2x3 is, therefore,


y = - y2y3(x - x2) + y2,

The clinant of lines perpendicular to x2x3 is then y2y3. The altitude through x1 is one of those perpendiculars and its equation is simply:


y = y2y3(x - x1) + y1

From (1)-(2) we easily find the foot of the altitude through x1. Just add up the two equations and pass to the conjugates:

x = (x1 + x2 + x2x3(y2 - y1))/2
= (x1 + x2 + x3 - x2x3/x1)/2
= (H - x2x3/x1)/2,

where H = x1 + x2 + x3. Similarly to (2), the altitudes through x2 and x3 are given by

y = y1y3(x - x2) + y2  and 
y = y1y2(x - x3) + y3

Multiply the first equation by y2 and the second by y3. Subtracting the results eliminates x. But solving for y and then changing to conjugates will resurrect x in an unexpectedly simple form: x = H! Since H depends symmetrically on the vertices x1, x2, and x3, all three altitudes meet at the point x = H, which proves existence of the orthocenter as promised.

It's time for the bonus. Note that, in our triangle,

  • the circumcenter O is at the origin: O = 0,
  • the orthocenter H is at the point x1 + x2 + x3,
  • the centroid M is known to be defined by (x1 + x2 + x3)/3.

This implies two things. First, the three centers - O, H, and M - lie on a straight line. And second, M is one third way from O to H. In other words,

  2 OM = MH,

The line is known as the Euler line or Euler's line.


  1. S. L. Greitzer, Conjugate Coordinate Geometry, Arbelos, v3, n1-5, MAA, 1997
  2. Liang-shin Hahn, Complex Numbers & Geometry, MAA, 1994

Related material

  • The Euler Line and the 9-Point Circle
  • Existence of the Euler Line: An Elementary Proof
  • Euler Line Cuts Off Equilateral Triangle
  • Fermat Points and Concurrent Euler Lines
  • Remarkable Line in Cyclic Quadrilateral
  • Nine Point Circle: an Elementary Proof
  • Three Euler Lines That Are Four
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