I am having trouble forming a figure for this problem.
The tangents at points $A$, $B$, $C$ at the circumcircle of $\triangle ABC$ meet the lines $BC$, $CA$, $AB$ at $M,N,P$. Show that $M,N,P$ are collinear.
when I draw the tangent lines to points $A, B, C$ they just form a bigger triangle surrounding the circle. So I am having trouble picturing this diagram. Any help would be appreciated.
If $O$ is the circumcenter of $ABC$, the tangent to the circumcircle at $A$ is just the perpendicular to $OA$ through $A$.
By Desargues' theorem the claim follows by proving that
On the other hand, the circumcenter of $ABC$ is the incenter of $A'B'C'$, hence the claim follows from the existence of the Gergonne point. The perspector of $ABC$ and $A'B'C'$ is also the symmedian point of $ABC$.