Here is one possible triangulation.
Note that all triangulations (that do not introduce extra vertices) will have the same number of triangles, which depends only on the number of sides in the polygon.
A sketch of the proof: A polygon with three sides is already a triangle, so the minimal number of triangles needed to triangulate it is exactly $1$. For a polygon with $n$ sides, choose two adjacent sides and connect the two vertices they do not share with a line. This creates a triangle and a new polygon. The new polygon is missing the two sides we chose, but has an extra side (the new side we just created). Thus the number of triangles needed to triangulate a polygon with $n$ sides is $1$ plus the number of triangles needed to triangulate a polygon with $n-2+1=n-1$ sides. As a recurrence relation:
a(3) &= 1 \\
a(n) &= 1 + a(n-1) \\
Solving this yields $a(n) = n-2$, so the number of triangles needed to triangulate a polygon with $n$ sides is exactly $n-2$, regardless of that polygon's shape.