Geometry question, triangles

Question

Alice chose a point $P_0$ on the side $AB$ of triangle $ABC$. Bob drew a circle centered at $B$ through $P_0$, which intersected side $BC$ in $P_1$. Then she drew a circle through $P_1$ centered at $C$ that intersected side $CA$ at $P_2$, then a circle centered at $A$ through $P_2$ which intersected side $AB$ at $P_3$. But Bob did not stop here – she drew another circle centered at $B$, this time going through $P_3$ to get $P_4$ on $BC$.

She continued in the same way, and all her points $P_0, P_1, P_2, \dots$ were on the sides. After a few more steps she noticed a pattern. What is this pattern? How would you explain it?

I have found the pattern ($P_6$ becomes $P_0$), but I cannot explain it!

Answer 1

thanks for your tips, here's my answer:

The pattern is that for any point along AB, the line goes back to the starting point.

Explanation:

I will first define some values.

a is the line opposite A (BC), b is the line opposite B (AC) and c is the line opposite C (AB).

$r_$ is the first radius from B to $P_{0_0}$. So:
$s_0=c-r_0$
$t_0=b-s_0=b-c+r_0$
$u_0=a-t_0=a-b+c-r_0$
$v_0=c-u_0=c-a+b-c+r_0=-a+b+r_0$
$w_0=b-v_0=b+a-b-r_0=a-r_0$
Now I will do the next point:

$x_0=a-w_0=r_0$

Since the seventh radius is indeed the first, I can confirm that the line always goes back to the starting point.