# 3-colourability of Eulerian maximal planar graph

The following paragraph is from this answer by David Eppstein (emphasis mine).

A maximal planar graph is 3-colorable iff it is Eulerian (if it is not Eulerian, then the odd wheel surrounding a single odd vertex requires four colors, and if it is Eulerian then a 3-coloring may be obtained by coloring a triangle and then extending the coloring in the obvious way to adjacent triangles).

(A maximal planar graph is a graph with a planar embedding such that every face is a triangle).

I don't quite understand how emphasized part works as a proof. Let $$G$$ be a maximal planar graph. If $$G$$ is 3-colourable, then the 3-colouring is unique (upto swapping of colours) because in every diamond subgraph of $$G$$, both degree-2 vertices of the subgraph should get the same colour. I suppose this suggests an algorithmic method for testing 3-colourability of $$G$$ namely (i) pick a triangle, (ii) give a 3-colouring to it arbitrarily, (iii) repeatedly extend it to adjacent triangles (assuming there is no inconsistency), and (iv) finally verify that the assignment is indeed a 3-colouring. If there is a vertex of odd degree, we will not succeed in assigning colour to all vertices. My question is this:- How can we guarantee that the assignment will be consistent provided all vertices have even degree?

Note: I don't see how Eulerian property (aka all vertices having even degree) ensures consistency of the assignment. I am asking this as a new question rather than a comment to the linked answer because he was answering a different question.