For planar triangulation, equivalence between 4-connectedness and non existence of separating triangle. I want to prove the following equivalence:
"A planar triangulation is 4-connected if and only if it has no separating triangle."
My attempts so far:
$\Rightarrow$:
If there is a separating triangle then by definition its removal separates the graph. This graph is therefore not 4-connected.
$\Leftarrow$:
Any triangulation is 3-connected: Considering one vertex, the subgraph induced by its neighborhood is $K_3$ which is 2-connected. We removed 2 vertices and the graph remains connected.
Now that we know the triangulation is 3-connected, we can use Whitney Theorem:"if $G$ is planar 3-connected, its faces in any planar represention are precisely its non-separating induced cycles." But I don't really see how.
Are my proofs right so far? Is it possible to conclude with Whitney theorem?
EDIT
A separating triangle is a cycle of length 3 whose removal separates the graph.
If there is no separating triangle then any cycle of length 3 is non-separating the graph.
Any triangulation is planar and 3-connected (I realize my proof of the 3-connectedness does not work, still working on it...), so we can use Whitney and claim that for any separating triangle, none bounds any face of the graph. Hence there is at least one vertex inside and outside of the separating triangle. 
I think the fact a plane triangulation is a maximal planar may also be useful.
But I cannot find a valid proof...
 A: "A planar triangulation is 4-connected if and only if it has no separating triangle."
$\Rightarrow$:
If there is a separating triangle then there is a 3-cut set. The graph is therefore not 4-connected.
$\Leftarrow$:
A planar triangulation is 3-connected.
If the graph is not 4-connected, then any minimal cutset $S$ is a set of 3 vertices.
A planar triangulation is a chordal graph. And in a chordal graph, any minimal cutset is a clique.
So $S$ is a separating triangle.
A: $\Rightarrow$ :
If there is a seperating triangle this triangle is also a 3-cutset. The graph is thus not 4-connected.
$\Leftarrow$ :
Any planar triangulation is 3-connected.
Suppose now that $T$ is not $4$-connected. Then there must be some $3$-cutset (since it is $3$-connected) let us denote the vertices of this cutset by $x, y$ and $z$. Removing this cutset splits the graph into at least two connected components $C_i$ and all components are incident to all cutvertices otherwise we would have found a $2$- or $1$-cutset.  
However, now $xy$ must be an edge in the triangulation $T$ otherwise the graph is not maximal planar (There can't be an edge incident to both $C_1$ and $C_2$ because that would negate $x, y ,z$ being a cutset.). In the same way $yz$ and $xz$ are edges of $T$. But then $xyz$ is a separating triangle. Hence this $3$-cutset can't exist and thus $T$ is $4$-connected
