There exist three consecutive vertices A, B, C in every convex n-gon with n≥3, such that the circumcircle of triangle ABC covers the whole n-gon From Problem Solving Strategies by Arthur Engel:
Problem to prove:
There exist three consecutive vertices $A$, $B$, $C$ in every convex $n$-gon with $n \ge 3$, such that the circumcircle of triangle $ABC$ covers the whole $n$-gon.
Proof by the author:
Among the finitely many circles through three vertices of the $n$-gon, there is a maximal circle.
Now we split the problem into $2$ parts:
(a) the maximal circle covers the $n$-gon.
(b) the maximal circle passes through three consecutive vertices.
We prove (a) indirectly. Suppose the point A' lies outside the maximal circle about triangle ABC where A, B, C are denoted such that A, B, C, A' are vertices of a convex quadrilateral. Then the circumcircle of triangle A'BC has a larger radius then that of triangle ABC. Contradiction.
We also prove (b) indirectly. Let A, B, C be vertices on the maximal circle, and let A' lie between B and C and not on the maximal circle. Because of (a), it lies inside that circle, but then the circle about triangle A'BC is larger than the maximal circumcircle. Contradiction.
Here is what I don't understand about the solution:
Is there sufficient proof for why the circumcircle of $A'BC$ has a larger radius than that of $ABC$? (for both the cases (a) and (b)).
Also, how is assuming the quadrilateral to be convex any helpful?
If I draw it out, I can kind of see that $A'BC$ has a larger radius than that of $ABC$, but it doesn't make sense intuitively how having points outside (part (a)) and inside (part (b)) of the circumcircle would both lead to larger radius circles?
Any help/suggestions would be much appreciated.
 A: I know this solution is a little late , but I hope it helps nevertheless.
Assuming the quadrilateral to be convex simply limits the number of cases . The assumption can be made , as the question states that the polygon is convex.

(a) 2 Cases arise :-

Case 1:- $\angle BAC $ is acute.
Consider $\angle BA’C $ . It lies outside the circle , and is thus smaller than $\angle BAC$ . Thus , $$ \frac{BC}{\sin \angle BA’C} > \frac{BC}{\sin\angle BAC} $$ This implies that the circumradius of $\triangle A’BC $ is greater .

Case 2:- $\angle BAC \geq 90 $
Since $A’$ lies outside the circle , $\angle CA’A$ < $\angle CBA$. Thus,
$$\frac{AC}{\sin \angle CA’A} > \frac{AC}{\sin \angle CBA} $$ which implies that the circumradius of $\triangle A’AC$ is greater than that of $\triangle ABC$ .
Thus , from the two cases , it is clear that we can find a triangle , with a circumradius bigger than the maximal circumradius , a contradiction.

(b) For proving this , we consider a general case . Note that $A’$ , being in between $B$ and $C$ , always lies in the minor segment formed by $BC$ , and $\angle BA’C$ is thus always obtuse.

Let the angle subtended by arc $BC$ at $O$ equal $\theta$ . Since $A’$ lies in the minor segment , $\angle BA’C$ > 180 - $ \frac{\theta}{2} $
$O’$ must lie outside the triangle (as it is obtuse), and on the perpendicular bisector $B’C’$ . Let us assume that the circumradius of $\triangle A’BC $ is smaller than that of $\triangle ABC $ . Then $O’$ must lie on $MO$ . $\angle BO’C$ is greater than $\theta$ , say = $\theta$ + y . We may let $\angle BO’A’$ equal y + x , and $\angle CO’A’$ equal $\theta$ - x . 
Since triangles $BO’A’$ and $CO’A’$ are isosceles , we get $$\angle BA’C  = 90 - \frac{x+y}{2} + 90 - \frac{\theta - x}{2} = 180 - \frac{\theta}{2} - \frac{y}{2} $$
Or $$ \angle BA’C < 180 - \frac{\theta}{2} $$ 
This contradiction implies that $O’$ must lie on ray $OC’$ , and thus the circumradius of $\triangle BA’C$ is greater than that of $\triangle ABC $
A: I think it is a good move to consider the largest circle incident with three of the given points, and then to prove (a) and (b) in turn.
But the verbal arguments  put forward by the author as an "indirect proof" of (a) and (b) can only serve as guidelines, and not as an actual proof. They would have to be substantiated with figures. It then would become obvious that several cases have to be considered; in particular whether the primary triangle $ABC$ contains the midpoint of the maximal circle in its interior or not.
