Notes
quadrilateral splitting a semi-circle solution

Quadrilateral Splitting a Semi-Circle

Solution by Cyclic Quadrilateral, Angle at the Centre is Twice the Angle at the Circumference, Angles in a Triangle, and Angles on a Straight Line

In the above diagram, quadrilateral $E B F C$ is cyclic and so angles $F \hat{B} E$ and $E \hat{C} F$ add up to $180^\circ$. Then as angles at a point on a straight line also add up to $180^\circ$, angle $A \hat{C} E$ is equal to angle $F \hat{B} E$, which is $b$.

As the blue arcs have the same length as the red, the red arc is half a semi-circle and so the angle $C \hat{O} D$ is $90^\circ$. Using the fact that the angle at the centre is twice the angle at the circumference, this establishes angle $C \hat{E} D$ as $45^\circ$.

Since angles in a triangle add up to $180^\circ$, from triangle $A C E$ then $a + b + 45^\circ = 180^\circ$ and so $a + b = 135^\circ$.

Solution by Invariance Principle

In this version of the diagram, angle $a$ is $90^\circ$ and angle $b$ is $45^\circ$ so their sum is $135^\circ$.

There is another version of the diagram in which the red arc is centrally placed in the semi-circle and where point $F$ coincides with point $B$. In this diagram, angle $a$ can be calculated as $45^\circ$, but it appears that angle $b$ is $67.5^\circ$ which does not give the expected answer. However, by taking $F$ to the same place as $B$, the angle $b$ should now be thought of as the tangential angle at $B$ which is $90^\circ$. As this is not a trivial step, this doesn’t fully qualify as a solution by invariance principle as the purpose of such solutions is to make the answer more obvious.