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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.

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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.