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

I know that the lower bound for finding the convex hull for a point set is also $\Omega(n \log n)$. 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?

Edit: By Triangulation I mean connecting all the points via non-crossing edges such that they form triangles: https://en.wikipedia.org/wiki/Point_set_triangulation
No, that's not a valid argument.  You'd have to describe how to find which part of the triangulation is the convex hull, in $o(n \log n)$ time, to make that a valid argument.