Angle of Intersection of Circles is Equal at Both Points

From ProofWiki
Jump to navigation Jump to search

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 $C'A = C'B$


\(\ds CA\) \(=\) \(\ds CB\) 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 Congruence
\(\ds \leadsto \ \ \) \(\ds \angle CAC'\) \(=\) \(\ds \angle CBC'\)

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

$\blacksquare$


Sources