Given any diagonal, consider triangulating the polygon on one side only of this diagonal. Clearly it must include at least one 'primitive' triangle formed of three consecutive vertices, and which is necessarily isosceles. So the triangulation of the polygon on both sides of the diagonal must include at least two such primitive isosceles triangles. Assume we have just two such primitive isosceles triangles, and denote the endpoints of one as P and Q (where Q is two vertices clockwise from P), and the endpoints of the other as R and S (where S is two vertices clockwise from R), and P,Q,R,S number clockwise round the polygon. Consider now triangulating the region between PQ and RS. To avoid generating any more primitive isosceles triangles, any further diagonals must 'separate' PQ and RS - i.e. each must have one vertex in the range Q to R inclusive (going clockwise) and the other vertex in the range S to P inclusive (going clockwise). Let us also denote the 'length' of a diagonal by the number of edges of the polygon that it subtends. So PQ and RS both have 'length' 2. To triangulate the region between PQ and RS, let us begin (WLOG) from P with a diagonal that runs from P to X where X is one vertex clockwise from Q. So PX has 'length' 3. The next diagonal can be from P to Y where Y is one vertex clockwise from X, or it can be from X to N where N is one vertex anti-clockwise from P. In either case, PY or XN have 'length' 4. Continue in this way, noting that the 'length' of each successive diagonal increases by one each time. Eventually we will get to a diagonal D with 'length' n, and then to diagonal E with 'length' n+1. But since there are 2n+1 edges of the polygon, D and E have the same physical length (E subtends n edges, counting the 'other way, via R, S). So D,E form our third isosceles triangle.

Denote the given polygon by $P$ and note that no triangle contains more than two sides of $P$. Moreover, since $P$ is regular, any triangle containing two sides of $P$ must be isosceles. Since there are $2n-1$ triangles and $2n+1$ sides of $P$, by Pigeonhole principle one of the triangles contains two of these sides. Excluding this triangle we are left with $2n-2$ triangles and $2n-1$ unused sides of the polygon. Again by Pigeonhole principle one more triangle contains two sides of $P$. Excluding these two triangles we are left with $2n-3$ triangles and $2n-3$ unused sides of the polygon. If there is one more triangle which contains two of these remaining sides, then we are done. Otherwise, each of the remaining triangles contains exactly one side of $P$. For each such triangle $ABC$, where $A, B, C$ located on $P$ clockwise and $AB$ is its side, let $x$ denote the number of sides of $P$ between $B$ and $C$ and let $y$ be the number of sides of $P$ between $A$ and $C$. With this way we assign a pair $(x, y)$ to each of the remaining triangles. Note that if $ABC$ and $A'B'C'$ are assigned the same pair, where $AB$ and $A'B'$ are sides of $P$, then either $BC$ intersects $B'C'$ or $AC$ intersects $A'C'$ (depending if $A'$ and $B'$ both are located between $B$ and $C$ or $A$ and $C$). This means pairs $(x, y)$ assigned to triangles are different. Moreover, we have $x+y=2n$. Since none of the remaining $2n-3$ triangles contains more than than one side of $P$ we have $x\neq 1$ and $y\neq 1$. Thus $$(x, y) \in \{(2, 2n-2), (3, 2n-3), \dots, (2n-2, 2)\}$$Since the set on the right hand side has exactly $2n-3$ elements, pairs assigned to triangles get all of these values. In particular, one of them is $(n, n)$ which means this triangle is also isosceles.