# Solution to the [[A Semi-Circle and a Quarter Circle]] Puzzle +-- {.image} [[ASemiCircleandaQuarterCircle.png:pic]] > The two blue angles are equal. What’s the size of the orange angle? =-- ## Solution by [[Angle in a Semi-Circle]], [[Angle at the Centre is Twice the Angle at the Circumference]], and [[Angles in a Triangle]] +-- {.image} [[ASemiCircleandaQuarterCircleLabelled.png:pic]] =-- With the points labelled as above, angle $B \hat{C} D$ is half of the reflex angle $B \hat{O} D$ since the [[angle at the circumference is half the angle at the centre]]. As it is the outside of a quarter circle, angle $B \hat{O} D$ is $270^\circ$ so angle $B \hat{C} D$ is $135^\circ$. Therefore, angles $B \hat{C} A$ and $A \hat{C} D$ add up to $135^\circ$. Since angles $A \hat{C} D$ and $A \hat{B} C$ are the same, this means that angles $A \hat{B} C$ and $B \hat{C} A$ add up to $135^\circ$. Then since the [[angles in a triangle]] add up to $180^\circ$, angle $B \hat{A} C$ is $45^\circ$. Finally, angle $O \hat{A} B$ is $90^\circ$ as it is the [[angle in a semi-circle]]. Hence angle $O \hat{A} C$ is $45^\circ$.