Charlie Conway 7/23/2009

**Pedal Triangle**

The **pedal triangle** RST of any triangle ABC is the triangle formed by connecting the three intersection points of lines perpendicular to AB, BC, and CA from the arbitrary **pedal point** P and the corresponding sides of ABC.

Click **here** for a workable sketch of the pedal triangle for triangle ABC.

What if the pedal point was not arbitrary? What observations could we make if we put the pedal point in a certain position relative to triangle ABC? Let's see what we can find by making the **circumcenter** the pedal point. The circumcenter is the intersection of the perpendicular bisectors of the three sides of the triangle.

**Observations**

1.) As seen top image shows, the arbitrary pedal point creates a pedal triangle that can be outside of the original triangle, save for one point on one side. Or the pedal triangle can be partially contained in the original triangle and partially outside. Or the arbitrary pedal point can lead to a pedal triangle fully contained in the original triangle. In contrast to the pedal triangle with an arbitrary pedal point, the pedal triangle with the circumcenter as the pedal point is *always*** inside** of the original triangle ABC.

2.) The pedal triangle TSR with pedal point P the circumcenter of ABC divides ABC into four congruent triangles, ARS, RCT, STB, TSR. To prove this, we will first show that the three blue triangles, ARS, RCT, and STB are congruent.

Without loss of generality, let's start with RCT and STB. CT and TB have equal length due to the fact that T is the midpoint of CB. Since S is the midpoint of AB and T is the midpoint of CB, then by the midpoint segment theorem ST is half the length of AC. Therefore ST has the same length as RC. Again, because R is the midpoint of AC and T is the midpoint of CB, RT is half of AB. Thus RT is the same length as SB. So by Side-Side-Side congruence, triangle RCT is congruent to triangle STB. And the same argument can be made to show that triangle ARS is also congruent to triangles RCT and STB.

The proof that triangle TSR is congruent to triangles ARS, RCT, and STB is similar. Without loss of generality, we will prove that TSR is congruent to RCT, and therefore is congruent to triangles ARS and STB. We have already shown that RS and CT are equal in length due to the midpoint segment theorem. In the same way, we have also shown that TS and RC have equal length. The two triangles share side RT. Therefore, triangle TSR is congruent to triangle RCT and is subsequently congruent to triangles ARS and STB.

3.)We can also show that triangles ARS, RCT, STB, and TSR are similar to triangle ABC. Without loss of generality, let's prove that triangle STB is similar to triangle ACB. SB is half of AB because is is the midpoint of AB. TB is half of CT because T is the midpoint of CB. And ST is half of AC due to the midpoint segment theorem. Thus by Side-Side-Side, triangle STB is similar to triangle ACB with a similarity ratio of 1:2. And therefore, triangles ARS, RCT, and TSR are also similar to triangle ACB with the same 1:2 ratio.