Prove a point is inside a circle through 3 other points This question builds upon a previous question I asked here
The answers helped me understand that for $4$ points that are not concyclic, and for which no $3$ fall on a straight line, of the $4$ circles that can be drawn through triples of the points, exactly $1$ or $2$ of the circles will contain the point they don't pass through.
I'm looking for an elementary proof (i.e. one that a high school geometry student could understand) as to why a point must be contained in a circle through $3$ given points. Consider the diagram below:

DLeMeur's answer to my original question helped me understand that circle $ABD$ will contain $C$ if and only if $D$ is placed in one of the gray areas. The arguments that I can make for this are only sort of convincing, but not really "air tight".
Case 1: $D$ is in the circular segment cut off by chord $\overline{AB}$. Then circle $ABD$ has a greater radius than circle $ABC$, and since $D$ and $C$ are on opposite sides of $\overleftrightarrow{AB}$, $C$ must be contained in circle $ABD$.
Case 2: $D$ is outside circle $ABC$, on the same side of $\overleftrightarrow{AB}$ as point $C$. Again, circle $ABD$ has greater radius than circle $ABC$, thus the entire portion of circle $ABC$ below $\overleftrightarrow{AB}$ is contained in circle $ABD$.
These arguments seem like they are missing some details. For instance, if someone asked, "How do you know circle $ABD$ has greater radius than circle $ABC$?" I do not have a good answer. I would appreciate any input!
 A: Let O be the center of the originial circle. Produce OD to cut the original circle at D'.

Follow the sizes of the marked angles and we will arrive at R > r.
Note-1. The numerical values of the angles are for comparisons
Note-2. If D is an interior point of $\triangle ABD'$, $\angle ADB \gt \angle AD'B$ is a fact.
A: In fact you argument is almost perfect, and you even do not need to ask (and answer) the question which circle has a larger radius. It suffices to know that two distinct circles can intersect in at most two points (Euclid,Elements, Book III, Proposition 10).
Let two distinct circles intersect in two points. Draw the straight line through the intersection points. The line splits the plane in two half-planes. Consider one of the half-planes. Due to the Proposition cited above a part of one circle will be completely inside the part of the other circle in this half-plane. And in the other half-plane the relation will be opposite, i.e. the circle which was "inside" in one half-plane will be "outside" in the other half-plane.
This suffices to prove your claim. Just pick a point $D$ in a corresponding region, draw the circle  $ABD$ and consider its relation to the circle $ABC$ in view of the above statement.
