# Three tangents meet opposite edges in collinear points

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.

• You can use Menelaus Theorem. If you can't solve, then I can find somethings for you. Have a good day. Apr 24, 2017 at 22:54
• @scarface can you explian how to use menalous ?? Apr 25, 2017 at 16:55
• I have added the proof with using Menelaus Theorem. Stay in peace. Apr 25, 2017 at 21:07
• Also note that in Collinearity Question using Menelaus' Theorem, DeltaScuti_Fomalhautb suggested a cut-the-knot webpage with a Menelaus-based proof for this.
– MvG
Aug 8, 2017 at 23:56

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.