A planar graph is triangulated if its faces are bounded by three edges. A triangle of a planar graph is a separating triangle if it does not form the boundary of a face. That is, a separating triangle has vertices both inside it and outside it; therefore its removal separates the graph.
Given planar graphs $H_1$ and $H_2$, a graph $G$ is a stitch of $H_1$ and $H_2$ if there is a planar drawing $ϕ$ of $G$ and there are proper subgraphs $G_1$ and $G_2$ of $G$ isomorphic to $H_1$ and $H_2$ respectively, such that $G_1 ∪ G_2 = G$ and $T := G_1 ∩ G_2$ is a facial triangle with respect to both $ϕ|G_1$ and $ϕ|G_2$.
Let $G$ be a triangulated planar graph with a separating triangle $T$. We assume that $G$ be the stitch of $G_1$ and $G_2$ where $T := G_1 ∩ G_2$.
My question is, if $G_1$ and $G_2$ are non-Hamiltonian, is $G$ non-Hamiltonian?
This question can be posed more strongly: If either $G_1$ or $G_2$ is non-Hamiltonian, is $G$ necessarily non-Hamiltonian?
Let $G_1$ and $G_2$ have more than four vertices.
Let $u_1, u_2, u_3$ be the vertices of the stitch triangle. Suppose the hamiltonian cycle goes $u_1 \rightarrow ... \rightarrow u_2 \rightarrow ... \rightarrow u_3 \rightarrow ... \rightarrow u_1$. This divides the cycles in 3 paths that are fully in $G_1$ or fully in $G_2$. WLOG two of them are in $G_1$ and third one is in $G_2$. Thus the path fully in $G_2$ must visit all vertices of $G_2$ except one of the triangle. We easily see that this path can be extended to an hamiltonian cycle of $G_2$. Same for $G_1$, the two paths must meet at a vertex of $T$, the two other extremities are joined by an edge of $T$, which gives an hamiltonian cycle.
By contrapositive, we just showed the stronger result that if $G_1$ OR $G_2$ is non-hamiltonian, then their stitching on a triangular face is also non-hamiltonian, and this regardless of the fact that $G_1$ and $G_2$ are triangulated.