# Solution to the Two Congruent Rectangles and a Semi-Circle Puzzle +-- {.image} [[TwoCongruentRectanglesandaSemiCircle.png:pic]] > The two rectangles are congruent. What’s the angle? =-- ## Solution by [[congruent|Congruent Polygons]] and [[triangle|Angles of Triangles]] +-- {.image} [[TwoCongruentRectanglesandaSemiCircleLabelled.png:pic]] =-- 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 $360^\circ$.