Please help me solve this problem.
Problem 2.27 (BAMO 2012/4). Given a segment AB in the plane, choose on it a point M different from A and B. Two equilateral triangles AMC and BMD in the plane are constructed on the same side of segment AB . The circumcircles of the two triangles intersect in point M and another point N.
(a) Prove that AD and BC pass through point N.
(b) Prove that no matter where one chooses the point M along segment AB, all lines MN will pass through some fixed point K in the plane.
I have problem in part b. I don't want to use coordinates. Can someone give a hint?
EDIT: Unfortunately this question has been put on hold. I am making this edit to explain myself. I had solved the first part of the problem when I asked here, but I wasn't able to solve the second problem. I don't remember how much progress I had made when I asked here. I didn't think of searching BAMO 2012 before asking here. After the advice from comments, I never asked questions directly from competitions again. This question is old so I can't add my progress anymore, I am sorry I didn't ask properly but I request you to re-open it. I'm one question away from a question ban. If I'm question banned, then I would like to use this edit and thank the mathematical community for answering my questions.
Part b: Draw tangents to circumscribed circles at points $A$ and $B$. These tangents interesect at some point $K$.
It's a known fact that the angle between a tangent and a chord is equal to the corresponding inscribed angle. Because of that:
$$\angle KAB=\angle KBA=60^\circ\tag{1}$$
It follows that triangle $KAB$ is equilateral so:
$$KA=KB\tag{2}$$
Therefore point $K$ has the same power with respect to both circumscribed circles. And because of that, it has to be on the radical axis $MN$. So all lines $MN$ pass through fixed point $K$ defined by (1) and (2).