Show that a convex polygon is contained within the largest circle determined by three consecutive vertices 
Given a convex $n$-gon. The circumcircle is constructed for every triple of consecutive vertices of the polygon. We get the $n$ circles. Select the circle with the largest radius. Prove that the circle contains the polygon.

My work so far:
$n=3 -$ triangle - obviously.
$n=4 -$
If $\angle B = \max \left\{A,B,C,D \right\}$ then $ABCD \in \omega_{ABC}$

$n\ge 5$. I need help here.
 A: The crux of the argument is contained in the following proposition.
Proposition 1 Let $p$ and $q$ be points in the plane, $H$ be one half-space separated by the line through $p$ and $q$, and $D$ be a disk with $p$ and $q$ on its boundary and centered in $H$. If point $r\in H$ forms a circumcircle with $p$ and $q$ that has a smaller radius than that of $D$, then $r\in D$.
See the configuration in this proposition in the figure below. Any circle through $p$ and $q$ has a center on the line orthogonal to the line through $p$ and $q$ and passing halfway between those points. For that circle to have a radius smaller than $D$ and be centered in $H$, the center must be on dotted segment between $c$ (the center of $D$) and the midpoint between $p$ and $q$. And we can see that the portion of any such circle that is in $H$ (the dotted circle in the figure) is contained in $D$.

Next, we establish that ``adding vertices'' to a polygon only increases the size of the largest circumradius formed by consecutive vertices.
Proposition 2 Let $P$ be a polygon with vertices in clockwise order $v_0$, ..., $v_{n-1}$ and let $P'$ be a polygon with vertices $v_0$, ..., $v_{n-1}$, $v'$ (where $v'$ is not necessarily listed in clockwise order). The largest circumradius of three consecutive vertices of $P'$ is at least as large as that for $P$.
Assume the largest circumradius for $P$ is formed by vertices $v_{i-1}$, $v_i$, $v_{i+1}$. If $v'$ does not occur between vertices $v_{i-1}$ and $v_{i+1}$, then those vertices remain consecutive and thus the largest circumradius associated with $P'$ is at least as large as $P$. So we focus on the case that  $v'$ causes $v_{i-1}$, $v_i$ and $v_{i+1}$ to no longer be consecutive. Without loss of generality, suppose $v'$ lies between $v_i$ and $v_{i+1}$. Now there are two cases each providing a larger circumradius of a new set of consecutive vertices:

*

*If $v'$ lies inside the circumcircle of $v_{i-1}$, $v_i$, $v_{i+1}$, then the circumradius of $v_i$, $v'$, $v_{i+1}$ is larger.

*If $v'$ lies outside the circumcircle of $v_{i-1}$, $v_i$, $v_{i+1}$, then the circumradius of $v_{i-1}$, $v_i$, $v'$ is larger.


Now we can proceed to the main result. Let $P$ be a convex polygon with vertices $v_0$, $v_1$, ..., $v_n$ ordered in clockwise order. Without loss of generality, assume that $v_0$, $v_1$, $v_2$ form the largest circumcircle of any three consecutive vertices of the polygon.
The proof follows by induction, evaluating the sequence of polygons $P_i$ with vertices $v_0$, ..., $v_i$. The base case holds trivially for a single triangle $P_2$. The inductive step involves adding a single new vertex $v_i$ to the polygon and checking that vertex lies in the circumcircle of $v_0$, $v_1$, $v_2$. Proposition 2 ensures that $v_0$, $v_1$, $v_2$ provides the largest circumradius for each polygon, i.e., the largest circumradius for $P_i$ cannot be realized by a different set of consecutive vertices.
We complete the result by applying the Proposition 1 with $p=v_0$, $q=v_1$, $r=v_i$, and $D$ being the circumcircle of $v_0$, $v_1$, $v_2$. Convexity of the polygon ensures the assumptions of the proposition. Specifically, $v_2$ and $r$ must lie in the same half-space and the center of $D$ lies in the same half-space as $v_2$.
