Notes
two congruent rectangles and a semi-circle solution

Solution to the Two Congruent Rectangles and a Semi-Circle Puzzle

Solution by Congruent Polygons and Angles of Triangles

The goal is to show that in the above diagram then point $O$ is the centre of the semi-circle. As the rectangles are congruent, their diagonals are the same length and so the lengths of $A D$ and $D B$ are the same. Triangles $O D A$ and $O D B$ are both right-angled triangles with two sides the same length. They are therefore congruent and so the lengths of $O A$ and $O B$ are the same (this can also be deduced from Pythagoras' theorem). This establishes $O$ as the midpoint of the diameter $A B$ and so as the centre of the circle.

Since $O$ is the centre, $O E$ has the same length as $O A$. As this is also the same length as $O B$ and the rectangles are congruent, the line segment $A E$ has the same length as $O A$ and $O E$. Triangle $O E A$ is therefore equilateral and so angle $E \hat{O} A$ is $60^\circ$. Angle $D \hat{O} E$ is then $90^\circ - 60^\circ = 30^\circ$. As the rectangles are congruent, sides $O D$ and $E D$ have the same length, and so triangle $O D E$ is isosceles. This establishes angle $O \hat{E} D$ as also $30^\circ$ and so $E \hat{D} O = 180^\circ - 60^\circ = 120^\circ$ since angles in a triangle add up to $180^\circ$. Finally, angle $C \hat{D} E = 360^\circ - 120^\circ - 90^\circ = 150^\circ$ since angles at a point add up to $360^\circ$.