# Solution to the [[Two Quarter Circles in a Rectangle]] Puzzle +-- {.image} [[TwoQuarterCirclesinaSquareII.png:pic]] > What’s the sum of these two angles? =-- ## Solution by [[Isosceles Triangles]], [[Angles in a Triangle]], and Properties of [[Circles]] +-- {.image} [[TwoQuarterCirclesinaRectangleAnnotated.png:pic]] =-- With the points labelled as above, the point where the quarter circles touch, $F$, lies on the line between their centres, so $B F A$ is a straight line and is a diagonal of the rectangle. As $A$ is the centre of the red quarter circle, triangle $A F E$ is [[isosceles]] and therefore angle $A \hat{F} E$ is the same as angle $A \hat{E} F$. Similarly, angle $B \hat{F} C$ is the same as angle $B \hat{C} F$. Since [[angles in a triangle]] add up to $180^\circ$, the angles in triangle $A F E$ and in triangle $B C F$ add up together to $360^\circ$. So $$ 2 A \hat{E} F + E \hat{A} F + 2 B \hat{C} F + C \hat{B} F = 360^\circ $$ Then also $A B D$ is a triangle and angle $A \hat{D} B$ is a right-angle, so $E \hat{A} F + C \hat{B} F = 90^\circ$, meaning that: $$ A \hat{E} F + B \hat{C} F = \frac{360^\circ - 90^\circ}{2} = 135^\circ $$