Assume that is possible. Choose an example that has the least number of vertices. Call a vertex interior if it is not one of the vertices of the square. We claim that each interior vertex has degree at least $6$. If $I$ is an interior vertex and has two edges, say $IA, IB$. Then there must be two edges connecting $A,B$ and $I$ must be inside the region formed by the two edges. Then we can remove $I$ and the extra edge connecting $AB$ while keeping the other. The parity of the degrees are unchanged and we have a planar graph with fewer vertices. If $I$ is an interior vertex and has four edges, say $IA,IB,IC,ID$ in clockwise order. $A$ has even degree, hence there is at least another edge $AO$ that is next to $AB$ in the counter-clockwise order. Then we have the triangle $AOB$. We remove $I$, remove $AB$, and add edges $OD,OC$. The parity of the degrees are unchanges and the new graph has fewer vertices. Thus each interior vertex has degree at least $6$, and at most one of the vertices of the square has degree $2$ (otherwise we would have just a square and one diagonal). A simple edge counting using Euler's formula shows that this planar graph has too many edges, impossible.

Here's another way of looking at this problem. Treat the triangulation as a planar graph. Take the dual graph of this planar graph $G^*$. The dual has every vertex of degree $3$ except one which has degree $4$, and every face is an even cycle. It is easy to prove that such a graph can't have an odd cycle (I can write the proof if anybody wants), so it is bipartite. However, one side of the bipartition has $1$ mod $3$ edges, and the other has $0$ mod $3$ edges, which is a contradiction.