Notes
two overlapping triangles and a circle solution

Solution to the Two Overlapping Triangles and a Circle Puzzle

Solution by Vertically Opposite Angles, Alternate Segment Theorem, and Angles in a Triangle

With the points labelled as above, angles $F \hat{H} C$ and $E \hat{H} B$ are equal as they are vertically opposite, then angles $F \hat{H} C$ and $H \hat{G} C$ are equal by the alternate segment theorem. So angle $H \hat{G} C$ is equal to angle $E \hat{H} B$. By a similar argument, angle $C \hat{H} G$ is equal to angle $A \hat{G} D$. Angles $C \hat{H} G$, $H \hat{G} C$, and $G \hat{C} H$ are the three angles interior to triangle $C G H$ and so sum to $180^\circ$.

Solution by Angles in a Quadrilateral, Angle at the Centre is Twice the Angle at the Circumference, Angle Between a Radius and Tangent

With the points labelled as above, let us write $a$ for angle $A \hat{G} D$, $b$ for angle $E \hat{H} B$, and $c$ for angle $G \hat{C} H$.

The anti-clockwise angle $G \hat{O} H$ is twice angle $G \hat{C} H$ as the angle at the centre is twice the angle at the circumference, so is $2 c$. So the anti-clockwise angle $H \hat{O} G$ is $360^\circ - 2 c$.

Angles $F \hat{H} O$ and $O \hat{G} F$ are $90^\circ$ as they are both the angles between a radius and tangent. So in quadrilateral $O H F G$, angle $G \hat{F} H$ is $180^\circ - (360^\circ - 2 c) = 2 c - 180^\circ$.

Angle $F \hat{H} C$ is equal to angle $E \hat{H} B$, so is $b$, as they are vertically opposite angles. Similarly, angle $C \hat{G} F$ is equal to $a$.

The angle sum of quadrilateral $G C H F$ is then $a + (360^\circ - c) + b + 2 c - 180^\circ = a + b + c + 180^\circ$. Since the sum of the angles in a quadrilateral is $360^\circ$, this shows that $a + b + c = 180^\circ$.

Solution by Invariance Principle

In the above configuration, triangle $D E F$ is equilateral. Angle $H \hat{O} G$ is $120^\circ$, so triangles $H O C$ and $C O G$ are also equilateral making angle $G \hat{C} H$ equal to $120^\circ$. Triangle $A B C$ is isosceles meaning that angles $H \hat{B} E$ and $D \hat{A} G$ are $30^\circ$. Angles $B \hat{E} H$ and $G \hat{D} A$ are $120^\circ$, meaning that angles $E \hat{H} B$ and $A \hat{G} D$ are $30^\circ$. So the marked angles sum to $120^\circ + 30^\circ + 30^\circ = 180^\circ$.