# 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 $CBC'$.

$CA = CB$ and $C'A = C'B$
 $\ds CA$ $=$ $\ds CB$ Definition of Radius of Circle $\ds C'A$ $=$ $\ds C'B$ Definition of Radius of Circle $\ds \leadsto \ \$ $\ds \triangle CAC'$ $=$ $\ds \triangle CBC'$ Triangle Side-Side-Side Equality $\ds \leadsto \ \$ $\ds \angle CAC'$ $=$ $\ds \angle CBC'$
$\blacksquare$