More specifically, consider a triangle ABC, and a point P that is not one of the vertices A, B, C. Drop perpendiculars from P to the three sides of the triangle (these may need to be produced, i.e., extended). Label L, M, N the intersections of the lines from P with the sides BC, AC, AB. The pedal triangle is then LMN.
The location of the chosen point P relative to the chosen triangle ABC gives rise to some special cases:
- If P = orthocenter, then LMN = orthic triangle.
- If P = incenter, then LMN = intouch triangle.
- If P = circumcenter, then LMN = medial triangle.
The vertices of the pedal triangle of an interior point P, as shown in the top diagram, divide the sides of the original triangle in such a way as to satisfy:85-86
If P has trilinear coordinates p : q : r, then the vertices L,M,N of the pedal triangle of P are given by
- L = 0 : q + p cos C : r + p cos B
- M = p + q cos C : 0 : r + q cos A
- N = p + r cos B : q + r cos A : 0
One vertex, L', of the antipedal triangle of P is the point of intersection of the perpendicular to BP through B and the perpendicular to CP through C. Its other vertices, M ' and N ', are constructed analogously. Trilinear coordinates are given by
- L' = − (q + p cos C)(r + p cos B) : (r + p cos B)(p + q cos C) : (q + p cos C)(p + r cos B)
- M' = (r + q cos A)(q + p cos C) : − (r + q cos A)(p + q cos C) : (p + q cos C)(q + r cos A)
- N' = (q + r cos A)(r + p cos B) : (p + r cos B)(r + q cos A) : − (p + r cos B)(q + r cos A)
For example, the excentral triangle is the antipedal triangle of the incenter.
Suppose that P does not lie on any of the extended sides BC, CA, AB, and let P−1 denote the isogonal conjugate of P. The pedal triangle of P is homothetic to the antipedal triangle of P−1. The homothetic center (which is a triangle center if and only if P is a triangle center) is the point given in trilinear coordinates by
- ap(p + q cos C)(p + r cos B) : bq(q + r cos A)(q + p cos C) : cr(r + p cos B)(r + q cos A).
The product of the areas of the pedal triangle of P and the antipedal triangle of P−1 equals the square of the area of triangle ABC.
- Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Geometry, Dover Publishing Co., second revised edition, 1996.