# Quadrilateral Splitting a Semi-Circle +-- {.image} [[QuadrilateralSplittingaSemiCircle.png:pic]] > In total, the blue arcs have the same length as the red. What’s the sum of these two angles? =-- ## 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]] +-- {.image} [[QuadrilateralSplittingaSemiCircleLabelled.png:pic]] =-- In the above diagram, quadrilateral $E B F C$ is [[cyclic quadrilateral|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]] +-- {.image} [[QuadrilateralSplittingaSemiCircleSpecial.png:pic]] =-- 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.