Notes
two circles in a rectangle solution

Two Circles in a Rectangle

Solution by Angle Between a Radius and Tangent, Isosceles Triangle, and Angles in a Quadrilateral

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 $135^\circ$.