I have two circular arcs $ABCF$ and $ADEF$ which have the same endpoints $A, F$ and 'contain' the points $B, C$ and $D, E$ in the specified order.
How can I prove that there cannot be a circular arc $BCDE$, i.e. one that starts in $B$, intersects $C$ before $D$ and ends in $E$?
If both arcs are minor circular arcs, I can show that $B$ and $E$ must lie on different sides of the line through $C$ and $D$, hence either $BCD$ would be clockwise and $CDE$ counterclockwise or the other way around. Either way they could not form a circular arc $BCDE$ together.
However the case where one (or both) are major circular arcs cause me trouble.
Edit 1: Please note that the circular arcs may not overlap.
Edit 2: To clarify – the 'drawing' whose (non)existence I want to prove contains circular arcs $AB, BC, CF, AD, DE, EF, BC, CD, DE$. None of these may overlap and they shall form larger circular arcs together: $ABCF$ is made of $AB, BC, CF$; $ADEF$ is made of $AD, DE, EF$; and $BCDE$ is made of $BC, CD, DE$.