Is Dual Graph of a Triangulation of a Polygon Tree?
Below is an example of a polygon with a single hole. The red area is the interior of the polygon, and the white triangle in the center is the hole (which lies outside the polygon). In the figure, I've also drawn a triangulation of this polygon. The dual of this triangulation is a cycle with $6$ vertices, and this is not a tree.

However, is this thing I've drawn here really a polygon? That depends on which definition of &quot;polygon&quot; you use. A polygon is often defined as an area bounded by a single polygonal chain that does not intersect itself. This type of polygon is also known as a simple polygon. Sometimes, we want to talk about generalizations of this simple polygon. A polygon with holes is a connected area bounded by several polygonal chains that do not intersect themselves or with each-other.
So, for this answer, I will define a polygon to be either a simple polygon or a polygon with holes. This means that above, I've shown an example of a polygon with a triangulation of which the dual graph is not a tree. This particular polygon is actually an example of something that holds more generally: the dual of a triangulation of a polygon is a tree if and only if the polygon is simple. (Proof idea: since a polygon is connected, the dual graph of the triangulation is also connected. Observe that the boundary of the polygon is disconnected if and only if the dual graph of the triangulation contains a cycle. Since a tree is defined as a connected acyclic graph, the claim follows.)