Notes
four triangles in a square solution

Solution to the Four Triangles in a Square Puzzle

Solution by Areas of Special Triangles and Angle Properties

In the above diagram, triangle $O D B$ is equilateral so line segments $B D$ and $O B$ are the same length.

Triangle $B D C$ is isosceles and right-angled. The perpendicular distance of $C$ to the side $B D$ is therefore half of the length of $B D$, so the area of this triangle is a quarter of the square of the length of $B D$.

Triangle $A O B$ is also isosceles. Angle $D \hat{O} A$ is a right-angle and angle $D \hat{O} B$ is the interior angle of an equilateral triangle, so is $60^\circ$. This leaves angle $B \hat{O} A$ as $30^\circ$. In the above diagram, $E$ is the point on $O B$ such that angle $O \hat{E} A$ is a right-angle, so angle $O \hat{E} A$ is $180^\circ - 90^\circ - 30^\circ = 60^\circ$ as angles in a triangle add up to $180^\circ$. Triangle $O E A$ is therefore half an equilateral triangle, meaning that $A E$ has half the length of $O A$, which is the same as $O B$ and $B D$. The area of triangle $A O B$ is therefore a quarter of the square of the length of $O B$.

Triangles $B D C$ and $A O B$ therefore have the same area and so the ratio of yellow to green in the design is $1 : 1$.