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What edges are not in a Gabriel graph, yet in a Delauney graph?

Submitted by: @import:stackexchange-cs··
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Problem

It is know that the Gabriel graph of a point set $P \subset \mathbb{R}^2$, $\mathcal{GG}(P)$ is a subset of the corresponding Delauney graph $\mathcal{DG}(P)$, i.e. $\mathcal{GG}(P) \subseteq\mathcal{DG}(P)$.

A Gabriel graph is defined as:

The Gabriel graph $\mathcal{GG}(P)$ of a point set $P \subset \mathbb{R}^2$ is defined as follows. The vertex set is $P$ and two vertices $p, q$ in $P$ are connected by an edge if and only if the interior of the disk with diameter $\overline{pq}$ is empty and $p, q$
are the only two points on its boundary. In particular, this implies that $\overline{pq}$ is also an edge in the Delaunay graph of P.

However, I am struggling to visualize or understand in what cases an edge $(p, q)$ would be in $\mathcal{DG}(P)$ but not in $\mathcal{GG}(P)$. Could someone give me an example? It seems to me as, by the empty-circle property of Voronoi edges, and their duality with the Delaunay graph, every edge in the Delaunay graph should satisfy the definition of a Gabriel graph. It is obvious that I am missing some edge case.

Solution

Ah, I think I figured it out, thanks to the help from Discrete Lizard.

See the following image of a Delaunay triangulation, where I've highlighted one example where the edges would be included in DT, but not GG.

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Context

StackExchange Computer Science Q#147973, answer score: 6

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