Here this is a solution :
I will sketch the reason why this is in fact a triangulation : Note that we must show that there are no two triangles $F_1,\ F_2$ sharing three vertices
Note that upper five triangles share one vertex. If $F_1$ is one of the five triangles and $F_1,\ F_2$ share three vertices, then $F_2$ is also one of the five.
Any two of six vertices on the five do not coincide.
Hence remaining last choice for $F_1$ is one of the remaining 4 triangles.
Note that $F_2$ is also one of the remaining 4 triangles. For convenience assume that $ (0,0),\ (0,1),\ (1,1),\ (1,0)
$ are vertices for a square. If $F_1$ contains $(0,0)$, then $F_1$ contains an interior point in the square Hence since $F_2$ must contain the interior point, $F_2$ contains $(0,0)$. But mid points of side of square in $F_1,\ F_2$ do not coincide.