How can I show that the lower bound for getting a triangulation from a point set is $\Omega(n \cdot log(n))$?

I know that the lower bound for finding the convex hull for a point set is also $\Omega(n \cdot log(n))$. Is it valid to argue that since the convex hull is also part of the triangulation it can't be done faster or else the lower bound for the hull would also be lower?