[[!redirects two circles solutions]] # Solution to the [[Two Circles]] Puzzle +-- {.image} [[TwoCircles.jpeg:pic]] > What’s the angle? =-- ## Solution by [[Angles in the Same Segment]], Opposite Angles in a [[Cyclic Quadrilateral]], and [[Angle at the Centre is Twice the Angle at the Circumference]] +-- {.image} [[TwoCirclesAnnotated.jpeg:pic]] =-- Label the points as above, with $O$ the centre of the blue circle. Angles $E \hat{O} A$ and $E \hat{D} A$ are equal since they are [[angles in the same segment]]. Then $A D E C$ is a [[cyclic quadrilateral]] so angles $A \hat{C} E$ and $E \hat{D} A$ add up to $180^\circ$. But also angles $A \hat{C} B$, $C \hat{B} A$, and $B \hat{A} C$ add up to $180^\circ$ since they are [[angles in a triangle]]. Therefore, angles $C \hat{B} A$ and $B \hat{A} C$ add up to angle $E \hat{O} A$. Then angle $E \hat{O} A$ is twice angle $C \hat{B} A$ since the [[angle at the centre is twice the angle at the circumference]]. Hence angles $C \hat{B} A$ and $B \hat{A} C$ add up to twice angle $C \hat{B} A$, so they must be equal. Thus angle $C \hat{B}A$ is $50^\circ$.