**Exploration 9:**

**Pedal Triangles**

John Watson

△A'B'C' is the pedal triangle of △ABC.

△A''B''C'' is the pedal triangle of the pedal triangle of △ABC.

△A'''B'''C''' is the pedal triangle of the pedal triangle of the pedal triangle of △ABC.

In this case, the pedal point (P) is inside the triangle.

Take a look at the relationship between the corresponding angles of △ABC, △A'B'C' and △A''B''C''...

As long as the Pedal Point, P is located inside △ABC, the three sets of corresponding angles add up to 180 degrees.

Let's see what happens to the pedal triangles as the pedal point (P) changes location...

What if the pedal point (P) is on the triangle?

As the pedal point (P) approaches a side of △ABC...

(i) Two vertices of △A''B''C'' collapse to the same point (which is one of the vertices of △A'B'C').

(ii) All three vertices of △A'''B'''C''' collapse to the same point (which is one of the vertices of △A'B'C').

All three angles of △ABC are congruent to the corresponding angles of △A'''B'''C''' regardless of the position of the pedal point (P). This suggests that the triangles are similar.

Here is an animation that shows what happens to the construction as the pedal point moves inside and outside of the triangle.

Here is a link to the script tool for these types of constructions.

What if ...

The pedal point P is the centroid of triangle ABC?

What if P is the incenter . . . ?

Then the pedal triangle has its vertices at the tangent points of the incircle. Prove that the three segments from the vertices of triangle ABC to the tangent points of the incircle on the opposite side are concurrent. This point of concurrency is the Gergonne Point of triangle ABC. The triangle formed by the points of tangency of the incircle is called the Intouch Triangle or the Contact Triangle.

What if . . .

P is the Orthocenter . . . ?

Prove that the Pedal triangle is the Orthic triangle of the original triangle. Even if the Orthocenter outside ABC?

Script tool link