Pedal triangle

If has trilinear coordinates , then the vertices of the pedal triangle of are given by \begin{array}{ccccccc} L &=& 0 &:& q+p\cos C &:& r+p\cos B \\[2pt] M &=& p+q\cos C &:& 0 &:& r+q\cos A \\[2pt] N &=& p+r\cos B &:& q+r\cos A &:& 0 \end{array} ==Antipedal triangle==

One vertex, , of the antipedal triangle of is the point of intersection of the perpendicular to through and the perpendicular to through . Its other vertices, and , are constructed analogously. Trilinear coordinates are given by \begin{array}{ccrcrcr} 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) \\[2pt] 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) \\[2pt] 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) \end{array} For example, the excentral triangle is the antipedal triangle of the incenter. Suppose that does not lie on any of the extended sides , and let denote the isogonal conjugate of . The pedal triangle of is homothetic to the antipedal triangle of . The homothetic center (which is a triangle center if and only if 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 and the antipedal triangle of equals the square of the area of . == Pedal circle ==

The pedal circle is defined as the circumcircle of the pedal triangle. Note that the pedal circle is not defined for points lying on the circumcircle of the triangle. Pedal circle of isogonal conjugates For any point not lying on the circumcircle of the triangle, it is known that and its isogonal conjugate have a common pedal circle, whose center is the midpoint of these two points. ==References==