Editing Pedal triangle (section)

==Antipedal triangle==

One vertex, {{mvar|L'}}, of the '''antipedal triangle''' of {{mvar|P}} is the point of intersection of the perpendicular to {{mvar|BP}} through {{mvar|B}} and the perpendicular to {{mvar|CP}} through {{mvar|C}}. Its other vertices, {{mvar|M'}} and {{mvar|N'}}, are constructed analogously.  [[Trilinear coordinates]] are given by
<math display=block>\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}</math>


For example, the [[excentral triangle]] is the antipedal triangle of the incenter.

Suppose that {{mvar|P}} does not lie on any of the extended sides {{mvar|BC, CA, AB}}, and let {{math|''P''<sup> −1</sup>}} denote the [[isogonal conjugate]] of {{mvar|P}}.  The pedal triangle of {{mvar|P}} is [[Homothetic transformation|homothetic]] to the antipedal triangle of {{math|''P''<sup> −1</sup>}}.  The homothetic center (which is a triangle center if and only if {{mvar|P}} is a triangle center) is the point given in [[trilinear coordinates]] by

<math display=block>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)</math>

The product of the areas of the pedal triangle of {{mvar|P}} and the antipedal triangle of {{math|''P''<sup> −1</sup>}} equals the square of the area of {{math|△''ABC''}}.