(Non)existence of circular arcs through set of points

Question

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 $E$?

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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.

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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 $BC, CD, DE$.

Answer 1

This answer calls figure 1 below "a quadrilateral $BCDE$" and figure 2 below "a quadrilateral $BCED$". The vertices of figure 1 are $B,C,D,E$ clockwise. On the other hand, the vertices of figure 2 are $B,C,E,D$ clockwise.

$\qquad\qquad\qquad$

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To prove by contradiction that there is no cirlce on which $B,C,D,E$ exist in this order, we suppose that there is such a circle.

Let us see, under the supposition, what conditions the four point $B,C,D,E$ in the figure given in the question has to satisfy.

Now, suppose that there is a circle on which $B,C,D,E$ exist in this order. (see the figure below. Note that the quadrilateral here is $BCDE$, not $BCED$.)

$\qquad\qquad\qquad$

Then, from the inscribed angle theorem, we get $$\angle{BCE}=\angle{BDE}\tag1$$ $$\angle{CBD}=\angle{CED}\tag2$$ $$\angle{CDB}=\angle{CEB}\tag3$$ $$\angle{DCE}=\angle{DBE}\tag4$$

These are the necessary conditions which the four points $B,C,D,E$ in the figure given in the question has to satisfy to have a circle on which $B,C,D,E$ exist in this order.

Now let us come back to the figure given in the question.

We now know that the four points $B,C,E,D$ in the figure given in the question have to satisfy $(1)(2)(3)(4)$.

So, from $(1)(2)$, we have that the quadrilateral $BCED$ (not $BCDE$) has to be a parallelogram.

And from $(3)(4)$, the four inner angels have to be the same and so we have that the quadrilateral $BCED$ (again, not $BCDE$) has to be a rectangle. (see the figure below. The figure below is the same as the figure given in the question except that we have that the quadrilateral $BCED$ has to be a rectangle.)

$\qquad\qquad$

Then, finally, we see that the order of the four points on the circumscribed circle of the rectangle $BCED$ has to be $B,C,E,D$.

This contradicts the supposition that there is a circle on which $B,C,D,E$ exist in this order.

It follows from this that there is no circle on which $B,C,D,E$ exist in this order. $\blacksquare$

Answer 2

Suppose we have the unit circle centered at the origin. Then we can assign points $A$, $B$, $C$, $D$, $E$, and $F$ have coordinates $$A = \left( \cos \theta_1, \sin \theta_1 \right), $$ $$B = \left( \cos \theta_2, \sin \theta_2 \right), $$ $$C = \left( \cos \theta_3, \sin \theta_3 \right), $$ $$D = \left( \cos \theta_4, \sin \theta_4 \right), $$ $$E = \left( \cos \theta_5, \sin \theta_5 \right), $$ $$F = \left( \cos \theta_6, \sin \theta_6 \right), $$ where $$ \theta_1 < \theta_4 < \theta_5 < \theta_6 < \theta_3 < \theta_2.$$ Does this sort of reasoning take you any closer?

Answer 3

There is only one circle circumscribed circle which have for center the intercection of the bissection of the triangle.

The intersection of the bissections of a triangle is inside the Area of a triangle so it can not belong to both triangles !