# Solution to the Angle in a Circle and Semi-Circle Puzzle +-- {.image} [[AngleinaCircleandSemiCircle.png:pic]] > What's the angle? =-- ## Solution by [[Angles in a Cyclic Quadrilateral]], [[Angle at the Centre is Twice the Angle at the Circumference]], [[Angles in a Triangle]], [[Angles at a Point on a Straight Line]], and [[Vertically Opposite Angles]] +-- {.image} [[AngleinaCircleandSemiCircleLabelled.png:pic]] =-- In the above diagram, the point labelled $O$ is the centre of the semi-circle. Let angle $O \hat{A} C$ be $a$ and let $D \hat{B} O$ be $b$. Since [[opposite angles in a cyclic quadrilateral]] add up to $180^\circ$, angles $O \hat{A} C$ and $C \hat{D} B$ add up to $180^\circ$, so angle $C\hat{D} B$ is $180^\circ - a$. Then as [[angles at a point on a straight line]] also add up to $180^\circ$, angle $F \hat{D} C$ and $C\hat{D} B$ add up to $180^\circ$, so $F \hat{D}C$ is equal to angle $a$. Using [[opposite angles in a cyclic quadrilateral]] again, angles $C \hat{E} F$ and $F \hat{D} C$ add up to $180^\circ$ meaning that angle $C \hat{E} F$ is $180^\circ - a$. A similar argument shows that angle $E \hat{F} D$ is $180^\circ - b$. The sum of angles $C \hat{E} F$ and $E \hat{F} D$ is therefore $360^\circ - a - b$. As triangle $O C A$ is [[isosceles]] and the [[angles in a triangle]] add up to $180^\circ$, angle $C \hat{O} A$ is $180^\circ - 2 a$. Using a similar argument gives angle $B \hat{O} D$ as $180^\circ - 2 b$. Therefore the reflex angle $C \hat{O} D$ is $540^\circ - 2a - 2 b$. Since the [[angle at the centre is twice the angle at the circumference]], the reflex angle $C \hat{O} D$ is $270^\circ - a - b$. Applying this again shows that this is twice each of the angles $C \hat{E} D$ and $C \hat{F} D$. So the sum of the angles $C \hat{E} D$ and $C \hat{F} D$ is angle $270^\circ - a - b$. The sum of angles $G \hat{E} F$ and $E \hat{F} G$ is then the difference between these two sums, so is $90^\circ$. Since the [[angles in a triangle]] add up to $180^\circ$, angle $F \hat{G} E$ is $90^\circ$. Therefore as [[vertically opposite angles]] are equal, angle $C \hat{G} D$ is $90^\circ$.