# 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

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: 