Incenter–Excenter Lemma (chicken-claw)

Dependencies: Central ∠, arc, chord are pairwise equivalent (arc = central angle), Inscribed angles on same arc are equal; angle in a semicircle is right (inscribed angles on the same arc are equal), Exterior angle equals the sum of two non-adjacent interior angles (an exterior angle of a triangle = sum of the two non-adjacent interior angles), Isoceles converse: equal base angles ⇒ equal sides (equal angles $\Rightarrow$ equal sides), Linear pair sums to 180° (a linear pair sums to $180^\circ$), Thales converse (right angle $\Rightarrow$ on the circle whose diameter is the hypotenuse), Three perpendicular bisectors meet (circumcenter) (the circumscribed circle of three points is unique).

Statement

Let $\triangle ABC$ be inscribed in $\odot O$, $I$ the incenter of $\triangle ABC$, and $I_a$ the excenter opposite $\angle A$ (i.e. the common point of the external bisectors of $\angle B$ and $\angle C$ with the internal bisector of $\angle A$). Extend the internal bisector of $\angle A$ to meet $\odot O$ again at $M$ (i.e. the second intersection of the bisector of $\angle BAC$ with $\odot O$; geometrically, $M$ is the midpoint of arc $BC$ not containing $A$).

Then the four points $B$, $C$, $I$, $I_a$ all lie on the same circle, with center $M$ and radius

In other words,

This is colloquially called the chicken-claw theorem — with $M$ at the "claw center," the four points $B$, $C$, $I$, $I_a$ resemble four equal-length claws.