Let x be a point with interior angle > 180°. If there is no such x then the shape is convex, and any two points can be connected to cut the shape in two. Otherwise start at x and draw a ray that is almost in line with one of the two incident edges. Let this ray intersect the polygon in a point y. If y is a vertex we are done, but y is probably a point on one of the sides. Slowly pivot the ray about x and let y slide along the outside of the polygon. If the ray suddenly crashes into an inlet, let z be the new point of intersection. Now xz is a segment that cuts the shape in two. If the ray does not run into an inlet, y eventually becomes an end point, and xy cuts the shape in two. Repeat this process for each piece, and then each of those pieces, until the shape is triangulated.
But how many triangles are involved? Assume the segment xy cuts the shape in two, and when each piece is fully triangulated, it has two more edges than triangles. Put the two pieces together and count edges and triangles. Let the first piece have k edges and the second piece have l edges. Since xy is not part of the original shape, it has k+l-2 edges. And it has k-2 + l-2 triangles. Thus the polygon has two more edges than triangles. By induction, an n sided polygon has n-2 triangles inside, and the interior angles sum to 180×(n-2).
