Notes
overlapping triangles in a circle solution

Solution to the Overlapping Triangles in a Circle Puzzle

Solution by Angle at the Centre is Twice the Angle at the Circumference, Isosceles and Equilateral Triangles

Let the points be labelled as above.

Consider triangle $A H C$. As $A H I B$ is a square and $H I C$ an equilateral triangle, $A H$ and $H C$ have the same length so triangle $A H C$ is isosceles. Angle $C \hat{H} A$ is $90^\circ + 60^\circ = 150^\circ$ so angle $A \hat{C} H$ is $15^\circ$. A similar argument shows that angle $I \hat{C} B$ is also $15^\circ$, so angle $B \hat{C} A$ is $30^\circ$. This is half of angle $B \hat{O} A$ so as triangle $B O A$ is equilateral, since the angle at the centre is twice the angle at the circumference, $O$ is the centre of the circle.

Triangle $O G D$ is then isosceles with angle $120^\circ$ at $O$ (as it is twice angle $G \hat{C} D$). This means that angle $O \hat{D} E$ is $30^\circ$ which is the same as angle $E \hat{O} D$ so triangle $E O D$ is isosceles and can be thought of as two half equilateral triangles aligned along their short sides. The length of $O D$ is therefore $\sqrt{3}$ times the length of $O E$.

Then $O D$ is the same length as $O A$, and of $A B$, and of $H I$. So the side length of equilateral triangle $H I C$ is $\sqrt{3}$ times the side length of $E F O$. The area of $H I C$ is therefore $3$ times the area of $E F O$, and so $E F O$ has area $8$.