On triangulations and "coverage" of circumcircles

Question

Let $P$ be a convex quadrilateral defined by four vertices $a$, $b$, $c$, and $d$. Suppose that the circumcircle of $\triangle abd$ contains $c$.* Let $D(\triangle abc)$ to denote the area enclosed by the circumcircle of $\triangle abc$. My claim is that

$$D(\triangle abc) \cup D(\triangle acd)\subseteq D(\triangle abd) \cup D(\triangle bcd).$$

Any tips for proving this statement? Below is an example—clearly the red area is contained within the blue. Here is an interactive example I made: https://www.geogebra.org/calculator/pzw75awc.

*In the context of Delaunay triangulations, $\triangle abd$ and $\triangle bcd$ do not satisfy the Delaunay condition. Edge $ac$ is an illegal edge, and the Delaunay triagulation of $P$ consists of $\triangle abc$ and $\triangle acd$.

Answer 1

It's sufficient to prove $D(\triangle abc) \subseteq D(\triangle abd) \cup D(\triangle bcd)$ by symmetry under permutation of the labels $b,d$.

Divide the circumcircle of $abc$ into three arcs: $\frown_{ab}$, $\frown_{bc}$, $\frown_{ac}$. The circumcircle of $bcd$ intersects the circumcircle of $abc$ at $b$ and $c$, giving rise to three possible cases:

1. The two circumcircles are equal, in which case we're done.
2. $(\frown_{ab} \cup \frown_{ac}) \subset D(\triangle bcd)$;
3. $\frown_{bc}\, \subset D(\triangle bcd)$.

To show that case 2 prevails over case 3 it suffices to show that $a \subset D(\triangle bcd)$, which should follow from your comments on Delaunay triangulations.

Similarly, $(\frown_{ac} \cup \frown_{bc}) \subset D(\triangle abd)$, so that the perimeter of $D(\triangle abc)$ is entirely in $D(\triangle abd) \cup D(\triangle bcd)$. Finally, circles are convex.