# Angle of Intersection of Circles is Equal at Both Points

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## 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 $CBC'$.

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 Equality | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \angle CAC'\) | \(=\) | \(\ds \angle CBC'\) |

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

$\blacksquare$

## Sources

- 1933: D.M.Y. Sommerville:
*Analytical Conics*(3rd ed.) ... (previous) ... (next): Chapter $\text {III}$. The Circle: $15$. Angle of intersection of two circles