ABC be any triangle. P is any point inside the triangle ABC. \( PA_1, PB_1, PC_1 \) be the perpendiculars dropped from P on the sides BC, CA and AB respectively. \( A_1 B_1 C_1\) constitutes a pedal triangle.

Also see

Drop perpendiculars from P on \( A_1 B_1, , B_1 C_1, C_1 A_1 \) at \( C_2, A_2, B_2 \) respectively. \( A_2 B_2 C_2 \) known as the second pedal triangle.

Finally, repeat the process to have the third pedal triangle \(A_3 B_3 C_3 \).

**Proposition (easy angle chasing): **The third pedal triangle is similar to the original triangle ( \( \Delta ABC \sim \Delta A_3 B_3 C_3 \) )

## Spiral Similarity

Notice that quadrilateral \( q_1 = P A_1 B C_1 \) is cyclic (why?). Rotate \( q_1 \) by \( 180^\circ \) and dilate it by a factor of \( \frac {1}{8} \). This spiral similarity sends the vertex B to \( B_3 \).

**Exercise 1:** Proof this using complex bashing or otherwise.

**Exercise 2:** Normalize by recreating the process in an equilateral triangle.

**Remark:** It is interesting to note that \( P A_1 B C_1 \) appears to be spirally similar to \( PB_2 C_1 A_2 \) and \( PC_3 A_2 B_3 \) but that does not happen.