Prove Line Circumscribed Circle; Incircle; Excircle

Question

How can I prove, that the circumscribed circle of a triangle does exactly cross the middle of the line that goes from the incenter of the incircle of the triangle to the excenter of the excircle of the triangle?

Answer 1

By angle chasing, you can prove the statement. Here are some steps to do so.

Let $ABC$ be a triangle with circumcircle $\Gamma$. Write $I$ and $I_a$, respectively, for the incenter and the excenter opposite to the vertex $A$ of the triangle $ABC$. Denote by $M_a$ the midpoint of the arc $BC$ of $\Gamma$ that does not contain $A$.

1. Show that $\angle IBI_a=\dfrac{\pi}{2}=\angle ICI_a$. As a consequence, $IBI_aC$ is a cyclic quadrilateral. If $\omega_a$ is the circumcircle of the quadrilateral $IBI_aC$, then $II_a$ is a diameter of $\omega_a$.

2. Check that $M_aB=M_aC$. This is easy.

3. Prove that $A$, $I$, $M_a$, and $I_a$ are collinear. This is also easy. (Recall that $I$ and $I_a$ are on the internal angular bisector of $\angle BAC$.)

4. Use (2) and (3) to verify that $M_aB=M_aI$. This implies that $M_a$ is the circumcenter of the triangle $IBC$. However, as $I$, $B$, $I_a$, and $C$ are concyclic, the claim follows.