Let triangle ABC be any triangle. If P is any point in the plane, then the triangle formed by constructing perpendiculars to the sides of ABC (extended if necessary) locate three points R, S, and T that are the intersections. Triangle RST is the Pedal Triangle for Pedal Point P.
Use my Pedal Triangle script tool!
Now here is an interesting thought that never occurred to me, the pedal triangle of the pedal triangle of the pedal triangle of a pedal point. Here, Let me just show you...
I am going to prove 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.
To do this let's first move the pedal point P onto the circumcenter of triangle ABC and make a few observations.
Notice, triangle XYZ is the medial triangle of triangle ABC because the perpendicular lines from the pedal point P through the sides of triangle ABC are actually perpendicular bisectors to these sides. But, P is also the orthocenter of triangle XYZ, because X, Y, and Z are the midpoints of the sides of triangle ABC, which makes triangle RST the orthic triangle of triangle XYZ.
Notice also that P is the incenter of triangle RST because the altitudes of triangle XYZ are the angle bisectors of RST.
Furthermore, the angle bisectors of triangle RST are the perpendicular bisectors of triangle A'B'C'. This makes point P the circumcenter of triangle A'B'C'. And the incircle of triangle RST is the circumcircle of A'B'C'. Too cool!
Claim 1: Triangle XYZ is the medial triangle of ABC.
Since point P is the circumcenter of triangle ABC, we know that the lines through P to the sides of ABC are perpendicular bisectors of those sides.
So, X, Y, and Z are the midpoints.
This implies that CY and XC are both 1/2 the length of CA and BC, respectively. Triangle ABC and triangle YXC share angle XCY. So, triangles ABC and YXC are similar by SAS similarity.
An analogous argument can be made for triangles ZBX and AZY.
Therefore, triangle XYZ is the medial triangle of ABC and therefore similar to ABC.
Claim 2: Let XYZ be a triangle with altitudes XR, YS, and ZT. These altitudes are the angle bisectors of the triangle RST.
Triangle YZE is similar to triangle YXR by AA similarity.
(YZ)(YR)=(YX)(AE) from the similarity of YZE and YXR.
Triangle YTR is similar to triangle YXZ by the similarities in the steps above. The same argument shows that the triangle SZR and triangle STX is similar to triangle YZX.
Angle SRZ = angle YRT.
Angle XRS = angle TRX.
Ray RX is the bisector of angle TRS. The same argument shows that ray TZ is the angle bisector of angle STR, and ray SY is the bisector of angle RST.
Now we know that point P is the the incenter of triangle RST. In order to find the incircle for triangle RST, we have to construct perpendicular lines from point P to the sides of RST. Well we did that when we constructed the pedal triangle A'B'C'. This implies that the angle bisectors are also the perpendicular bisectors of the segments of triangle A'B'C'.
This implies that point P is the circumcenter of triangle A'B'C'. Since point P is also the circumcenter of triangle ABC, then triangle A'B'C' must be similar to triangle ABC. This completes the proof.
Have fun with my construction of the pedal triangle of the pedal triangle of the pedal triangle.