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The tangents at points $A$, $B$, and $C$ to the circumcircle of $\triangle ABC$ meet the lines $BC$, $CA$ and $AB$ at $M$, $N$, $P$.
Show that $M$, $N$ and $P$ are collinear.
I know I have to use radical axis but I just don't know where to start or how to draw it.
What makes you say “I know I have to use radical axis”?
Personally I'd treat this as a special case of Pascal's Theorem, if you can build on that:
Pascal's theorem states that given all the other incidences, the three points $M,N,P$ will be collinear in the above situation. Now you can imagine moving $A_1$ and $A_2$ closer together. In the limit, when they are the same point, the line connecting them will have become a tangent. Likewise for the other two. So all the dark blue lines would become tangents, all the cyan ones are opposite connecting lines. It doesn't even have to be a circle, any conic will do.
Also this problem is a special case of homothety centres and homothety line. Three of any homothetic figures are given. In pairs homothety centers of these are collinear. Especially, three circles are homothetic in pairs. Intersections of tangent lines are homothety centres. So, they're collinear.
Furthermore, OP well known Monge Theorem. (Link including a proof of it).
More and moreover, in Challenging Problems in Geometry (Alfred S. Posamentier, Charles T. Salkind) at pages 189-190:
