Isogonal Conjugates

The lines joining P to the three vertices are called the cevians of that point.

The Isogonal Conjugates of a point, P, is a point P' at the intersection of the cevians you get by reflecting each cevian about the bisector of the angle of the original triangle.

The orthocenter (the common intersection of the three altitudes of a triangle, or their extensions, which must meet in a point) and the circumcenter (the center of the circumscribing circle) are isogonal conjugates of one another.

First of all, it's not at all obvious that the reflected cevians will intersect in a single point.  So I'll first show that's true, and then draw a picture of the relationship between P and P'.

Let D, E, F be the feet of the cevians, as in the diagram.

Now, there is a well-known theorem called Ceva's theorem that states that given three arbitrary cevians AD, BE and CF, the three of them all meet at a point P if and only if

(AF)/(FB) (BD)/(DC) (CE)/(EA) = 1

Now, let's express this in terms of the angles x, y, and z as shown on the diagram.

We see that

(BD)/(DC)
= (area BAD) / (area DAC) as the two triangles have the same height
= (1/2 AB AD sin x)/(1/2 AD AC sin (A-x))
= (AB sin x)/(AC sin (A-x))

Similarly,

AF/FB = (AC sin z)/(BC sin (C-z)), and
CE/EA = (BC sin y)/(AB sin (A-y))

Hence the Ceva condition is that the lines intersect if and only if

(AB sin x)/(AC sin (A-x)) (AC sin z)/(BC sin (C-z)) (BC sin y)/(AB sin (A-y)) = 1

Now, all the side lengths suddenly cancel from this fraction, and we are left with

sin x sin y sin z = sin(A-x) sin (B-y) sin (C-z)

If we replace x, y and z in the above equation with A-x, B-y, C-z, then it clearly remains true.

That is, if you have a set of cevians concurring in a point P, the new set of cevians you get by reflecting each cevian about the bisector of the angle of the original triangle will also concur at a new point P'.

Related pages in this website:

Triangles -- a home page for anything about triangles.

Ceva's Theorem -- the relationship of the portions of the sides of a triangle cut by cevians

Proof that the median and altitude of a right triangle are reflections about the bisector iff ABC is a right triangle.

Triangle Trisection -- If a point, P, on the median of triangle ABC is the isogonal conjugate of point Q, on the altitude of ABC, then ABC is a right triangle.

The Orthocenter and Circumcenter of a triangle are isogonal conjugates.

The webmaster and author of the Math Help site is Graeme McRae.
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