Let $ABC$ any triangle. Consider $C_1$ the incircle and $C_2$ one of the three excircles.
The question is, what is the radical axis? That is the only information we have.
Working a lot with this problem, if we consider $A', B', C'$ the middle points of $BC, AC, AB$, respectively, we have that the radical axis of $C_1$ and $C_2$ is one external bisector of triangle $A'B'C'$.
How to prove this? Remembering that any point $P$ in the radical axis of $C_1$ and $C_2$ verify that $PO_1^2-r_1^2=PO_2^2-r_2^2$, where $O_1, O_2$ and $r_1, r_2$ are centers and radios of $C_1, C_2$, respectively.
This is abstract geometry, so we are not able to use results of analytic geometry.
Help! 🙁
Best Answer
The radical axis of two circles possesses the following properties.
(1) It is perpendicular to the line of centers; and (2) It bisects their direct common tangents.
Therefore, to find the radical axis of the in-circle and the corresponding ex-circle (wrt A), we do:-
(1) Locate P, the point of contact of the in-circle to the line BC.
(2) Locate Q, the point of contact of the ex-circle to the line BC.
(3) Locate M, the midpoint of PQ.
(4) Through M, drop…..