# Two Circles in a Rectangle +-- {.image} [[TwoCirclesinaRectangle.png:pic]] > What's the angle? =-- ## Solution by [[Angle Between a Radius and Tangent]], [[Isosceles Triangle]], and [[Angles in a Quadrilateral]] +-- {.image} [[TwoCirclesinaRectangleLabelled.png:pic]] =-- With the points labelled as above, point $O$ is the centre of the yellow circle. Angle $D \hat{C} O$ is the [[angle between a radius and tangent]] so is $90^\circ$, as is angle $O \hat{B} D$. Since triangle $C O B$ has two radii of the yellow circle, it is [[isosceles]] and so angles $B \hat{C} O$ and $O \hat{B} C$ are equal. Therefore, angles $D \hat{C} B$ and $C \hat{B} D$ are equal. (This can also be seen by considering how the diagram behaves under a [[reflection]] through the line through $O D$.) A similar argument shows that angles $B \hat{A} E$ and $D \hat{B} A$ are equal. Therefore, the sum of angles $B \hat{A} E$ and $D \hat{C} B$ is the same as angle $C \hat{B} A$. Adding up the angles in the quadrilateral $E A B C$, therefore, gives $90^\circ$ plus twice angle $C \hat{B} A$. Since the [[angles in a quadrilateral]] add up to $360^\circ$, angle $C \hat{B} A$ must be $135^\circ$.