The discussion of heuristics includes only information about admissibility and not consistency. An admissible but inconsistent heuristic can also cause A* to miss the correct solution.
In other words, if the heuristic is consistent, then it implies that A* will expand the optimal number of states.
Edit: The paper you cited uses a traversal slightly different from traditional A*.
So if you do A* with an inconsistent heuristic, you need to revisit nodes if you explore them a second time with a cheaper cost (i.e., you can re-expand nodes in your closed list). If you do this, you will find optimal solutions even with an inconsistent heuristic.
The only requirement on your heuristic if you A* to find optimal solutions is that it be admissible.
A nitpick: "breadth-first heuristic search" is an algorithm developed by Zhou and Hansen and, while it's related to A* (in that it uses an admissible heuristic to prune the search space), it's not actually a variant of A*. (It's not a best-first search algorithm, since it doesn't expand nodes in increasing order of f cost.)
http://web.cs.du.edu/~sturtevant/papers/incnew.pdf
Specifically, look at the Martelli G family example.
The numbers in the nodes represent heuristic values at each node and the numbers on edges represent edge weights; we're looking to get from S to T.
Our heuristic is admissible because it never overestimates the distance to the target.
Consistency dictates that the heuristic value at any node may at most be the weight of any out-edge added to the heuristic value of the node that the edge connects to. So in our example, the heuristic at A is 3, but at B it is 0, but 3 > 1 + 0, so our heuristic is inconsistent.
The reason this causes issues is because our algorithm assumes that any visited path is the shortest path to the end vertex of that path. So when we traverse this graph, we will visit B first, because the sum of its heuristic value and the respective edge weight is less than that of A.