# What is the radical axis of an incircle and excircle of an arbitrary triangle?

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! :(

• btw Gauss, how may the radical triangle be related to ABC? Commented Oct 4, 2020 at 10:36

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…..

• One can show that $\overline{BP}\cong\overline{CQ}$, making $M$ even easier to construct (and the target radical axis even easier to describe).
– Blue
Commented Oct 4, 2020 at 12:59
• @Blue A good point.
– Mick
Commented Oct 4, 2020 at 13:38