Angle of Intersection of Circles is Equal at Both Points

Theorem
Let $\CC$ and $\CC'$ be circles whose centers are at $C$ and $C'$ respectively.

Let $\CC$ and $\CC'$ intersect at $A$ and $B$.

The angle of intersection of $\CC$ and $\CC'$ at $A$ is equal to the angle of intersection of $\CC$ and $\CC'$ at $B$.

Proof
Consider the two triangles $CAC'$ and $CBC'$.


 * Circles-angle-intersection.png

By definition of radius:
 * $CA = CB$ and $C'A = C'B$

The result follows from Angle of Intersection of Circles equals Angle between Radii.