From point C the two tangent lines' touch points with circle k1 are A and B. Circle k2 touches the AB line at B(so if O2 is the centre f k2, then O2BA∠=90°), and C is on k2 too. The other crosspoint of the two circles is M. Prove that the AM line cuts BC into two equally long halves.
I know that BC needs to be equal to AC because AC and BC are tangents of k1, but I can't find out anything about M. I wanted to prove that, M is a point of the median, obviously, and I think this is the key though, but I don't know how to continue.
Also, I noticed that, if I would mirror circle $k_2$ trough BC, that I get is the circle around ABC triangle. So we can reframe the problem as: There are two circles($k_1$ and $k_2$) whit two points($A$ and $B$) of intersection. Prove that, if $k_1$ would be rotated around $B$ point(so we get $k_3$ circle), then $P = k_3 \cap k_2$ and $P \neq B$, AP line cuts the line connecting the two crosspoints of $k_1$ and $k_2$ into two equally long halves.