by Donny Thurston
1. Prove the pedal triangle of the pedal triangle of the pedal triangle of a point is similar to the original triangle. That is, show that the pedal triangle A'B'C' of pedal triangle RST of the pedal triangle XYZ of pedal point P is similar to triangle ABC.
In this write-up we are going to try and prove this fundamental, but sometimes tough to follow, theorem about pedal triangles. We will see how the result comes naturally when seeing how the first pedal triangle is related to the original triangle is some ways that aren't always so obvious. However, once we can see how the first two triangles are related, the inevitability of similarity between two triangles, eventually, becomes obvious.
When it comes to proving similarity, one of the easiest ways is often to examine the angles in the triangles. Similarity can also be proven by discovering proportional side lengths, but recall that all an inquiring mind has to discover about two triangles is that two sets of angles are congruent for two triangles to be similar, so that is our goal.
As a brief review, recall that pedal triangles are formed by choosing an arbitrary point inside or outside of a triangle (for this proof, we will be using a point inside), and then drawing the perpendicular lines that join that point to the sides (extended if necessary) to the point. The intersections of these lines with the sides of the original triangle form the pedal triangle. Observe:
Here we have pedal point P, and pedal triangle DEF to triangle ABC. We could draw more triangles (on our way to the third), but let us see if we can deduce any relationships here in this triangle. Since our goal is to be able to compare the angles of the first and third pedal triangle, let us consider the angles in this triangle.
Note that a circle can be draw connecting P, F, A, and D. These points all lie on the same circle because angle PFD and angle PDA subtend the same segment (PA) and are both right angles. Remember that, in right triangles, the hypotenuse is always the diameter of its circumcircle (the right angle represents an inscribed angle that must subtend a 180 degree angle, since 180 is 90*2). Since F and D share the same diameter for their circumscribes, then they must be on the same circle. See below:
Seeing this, we can come to another conclusion. Angle PDF must be congruent to PAF, since they subtend the same arc. This is because all inscribed angles are 1/2 of the measure of their subtended arc, and if two angles are half of the same measure, they must be equal. Lets call that angle, 1. Using similar reasoning we can decide that angle PFD and PAD are equal as well. We'll call that 2.
Using similar reasoning for each angle in the triangle, we can come see these relationships:
So, all of the angles are still there, but split apart and "jumbled" between the angles of the pedal triangle. However, the really interesting part comes when we create the next pedal triangle
Notice how the red dashed lines serve the same purpose in the first pedal triangle as the green dashed lines did in the original triangle. In fact, we can perform the exact same comparison as we did in the last part of this proof. For instance, P, G, H, and F are also on a circle, for the same reasons as before, and their angles can also be represented in the same portions of the original angles. Observe:
We see here that the angles (half-angles), while still intact, are mixed up again. However for the next pedal triangle, compared in the same way with circles built within the triangle creates a familiar set of angles:
When looking at this last figure we can observe, CAB = 1 + 2 = LKJ, CBA = 3 + 4 = JLK, ACB = 6 + 5 = LJK.
Therefore, triangle ABC is similar to triangle KLJ (the third pedal triangle), by the Angle-Angle Postulate.
