see below for details
Please enable Java for an interactive construction (with Cinderella).

The picture shows a monotone triangulation. The graph contains two monotone chains, the vertices of which are:
CHAIN 1 = {L, M, N, O, P, Q, A, C, D}
CHAIN 2 = {D, E, F, G, H, K, L}

The green vertices represent "special" cases.
Vertex N represents case 3 where O is vertex i and N is i-1. Both are on the same chain but the last diagonal LN does not form a convex vertex. Hence, N would go to the stack. The yellow egde represents the diagonal which would eventually be drawn.

Vertices F & A represent case 2 (both different instances of case 2). Vertex A and Q are both on the same polygonal chain and are adjacent in the sorted 'y' list. We look at edge HQ and see that the vertex there is convex. Hence we can draw a diagonal from H to A.

The same holds for vertex F. Its previous vertex G lies on the same polygonal chain. Hence we look at edge GA and see that the vertex formed is convex and so, we draw a diagonal from F to A.

All black diagonals represent case 1: Current vertex and previous are on separate chains.

Created with Cinderella