Arc Midpoints

Suppose we have a triangle $ABC$ and its circumcircle $\Gamma$. Let's construct the angle bisector of say, angle $A$.

Right, $M$ is the midpoint of arc $BC$. This is because $\angle BAM=\angle MAC$. Then, it follows that the arc lengths of $BM$ and $MC$ are equal. Thus, $M$ is the midpoint.

There are other interesting properties of $M$ too!

We also know that $M$ is on the perpendicular bisector of $BC$. This is because $M$ is equidistant from $B$ and $C$. This gives us an interesting property! The angle bisector of $A$ and perpendicular bisector of $BC$ intersect on the circumcircle! This is a very useful point in olympiad geometry.

Another useful fact is that if $I$ and $E$ are the incenter and $A$-excenter respectively, then $BICE$ is a cyclic quadrilateral, with $M$ as its center! Try to prove this one yourself.