**Kristin Ottofy**

Assignment 9

The pedal triangle of a pedal triangle of a pedal triangle is similar to the original triangle.

Proof: Let ABC be the original triangle. The pedal triangle of ABC is triangle XYZ. The pedal triangle of XYZ is triangle RST. The pedal triangle of RST is A'B'C' which is the triangle we are trying to prove is similar to ABC.

Here is a picture of the triangles and a GSP file.

I want to prove that angle ABC is similar to angle A'B'C' and so on.

Let point E be the midpoint of line segment BP, then let circle E be the circle at center E with radius EB.

I know that <PXB is congruent to <PZB and they are both right angles. Points X and Z subtend point P (and triangles PXB is congruent to triangle PZB). Therefore, B, X, Z, and P lie on the same circle. Now I can take line segment PZ as the base for triangle PBZ and triangle PZX where X and B are the third points. Because X and B both lie on circle E and have the same base PZ, angles ZBP and ZXP are congruent.

I can repeat this same process on the subsequent pedal triangles and discover that the angles PBZ = PBC, PXZ = PXR, PSR, and PB'C' are congruent. This process also works for angle PBX.

Here is a zoomed in version:

Finally, I can repeat this process on angles BCA and CAB to find that angle BCA is congruent to B'C'A' and CAB is congruent to C'A'B'.