Notes
multiple angles solution

Solution to the Multiple Angles Puzzle

Solution by Angles in a Triangle

The angles at $E$ can be filled in using angles at a point on a line add up to $180^\circ$, so angle $D \hat{E} A = 110^\circ$, $A \hat{E} O = 70^\circ$, and $O \hat{E} C = 110^\circ$.

Using the fact that angles in a triangle add up to $180^\circ$, various other angles can be now be filled in.

From triangle $A O D$, angle $O \hat{A} D$ is $50^\circ$. This means that triangle $A O D$ is isosceles, and so the length of $O D$ is the same as that of $O A$.

From triangle $A E O$, angle $O \hat{A} E$ is $30^\circ$. Then from triangle $A C B$, angle $A \hat{C} B$ is $90^\circ$.

The point $O$ is the midpoint of $A B$, so as triangle $A C B$ is a right-angled triangle, joining $O$ to $C$ results in two isosceles triangles. That is, $O C$ has the same length as $O A$ and $O B$.

Since triangle $C O B$ is isosceles, and angle $O \hat{B} C$ is $60^\circ$, triangle $C O B$ is actually equilateral. This means that angle $B \hat{O} C$ is $60^\circ$. So angle $C \hat{O} D$ is $40^\circ$ as angles at a point on a line add up to $180^\circ$.

Both $O D$ and $O C$ are equal to $O A$, so triangle $D O C$ is also isosceles. As angle $C \hat{O} D$ is $40^\circ$, angle $O \hat{D} C$ is $70^\circ$.

Finally, from triangle $D E C$, angle $D \hat{C} E$ is $40^\circ$.

Solution by Angle in a Semi-Circle and Angles in the Same Segment

In the above diagram, the semi-circle is centred on $O$ with diameter $A B$.

Angle $O \hat{A} D$ is $180^\circ - 80^\circ - 50^\circ = 50^\circ$ since angles in a triangle add up to $180^\circ$, so triangle $A O D$ is isosceles. Therefore $O D$ is the same length as $O A$ and so $D$ lies on the semi-circle.

Angle $A \hat{E} O$ is $70^\circ$ since vertically opposite angles are equal, so angle $O \hat{A} E$ is $180^\circ - 80^\circ - 70^\circ = 30^\circ$. Angle $A \hat{C} B$ is therefore $90^\circ$ and so as angle in a semi-circle is $90^\circ$, $C$ is also on the semi-circle.

Angles $D \hat{C} A$ and $D \hat{B} A$ are equal as they are angles in the same segment. Triangle $A D B$ is a right-angled triangle and angle $O \hat{A} D$ is $50^\circ$ so angle $D \hat{B} A = 40^\circ$. Hence the green angle is $40^\circ$.