# A Double Meaning of an Arc's Midpoint

Let $M$ be the midpoint of an arc of the circumcircle of $\Delta ABC$ apposite vertex $A$; let $I$ denote the incenter of the triangle. Then $M$, $I$, and $A$ are collinear. In addition, $CM=IM$. The applet below serves a dynamic illustration

Proof Let $M$ be the midpoint of an arc of the circumcircle of $\Delta ABC$ apposite vertex $A$; let $I$ denote the incenter of the triangle. Then $M$, $I$, and $A$ are collinear. In addition, $CM=IM$. ### Proof

The proposition is a consequence of the definitions. The Incenter lies at the intersection of angle bisectors; in particular, it belongs to the angle bisector from $A$.

On the other hand, equal inscribed angles are subtended by equal arcs, such that $\angle BAI=\angle CAI$ implies that the (second) intersection of $AI$ with the circumcircle is exactly the midpoint of the opposite arc $\stackrel \frown {BC}$, which is $M$.

Let angles at $A$ and $C$ be equal $2\alpha$ and $2\gamma$, respectively. Then, as an external angle of $\Delta AIC$, $\angle MIC=\alpha +\gamma$. But since inscribed angles $BAM$ and $BCM$ are both equal to $\alpha$, $\angle ICM=\alpha+\gamma$, implying $\angle ICM=\angle MIC$ and making $\Delta MIC$ isosceles, with $CM=IM$.

It is worth noting that the excenters of a triangle have a similar property:

Let $M$ be the midpoint of an arc of the circumcircle of $\Delta ABC$ apposite vertex $A$; let $I_A$ denote the excenter of the triangle opposite A. Then $M$, $I_A$, and $A$ are collinear. In addition, $CM=I_{A}M$. For one, $\Delta ICI_A$ is right so that the median from $C$ equals half the hypotenuse $II_A$. Point $M$ that forms an isosceles triangle $CMI$ is unique on the hypotenuse (because the base angle $MIC$ does not depend on the position of $M$.) It follows that $M$ is the midpoint of $II_A$ which proves the claimed property of the excenters.

Note that the above supplies an alternative proof of the previously established fact that the midpoint of $II_A$ lies on the circumcircle. As we've seen, this same point also serves as the midpoint of the arc of the circumcircle that is crossed by $II_A$. 