# Do divergent parts cancel out between the 1-loop contribution to the vertex and the self-energy on the electron legs of the vertex diagram?

I regret I don't know how to draw Feynman diagrams here, so I refer to the standard book Peskin-Schroeder. At the beginning of section 7.5 on page 244 of this book several Feynman diagrams are shown, I will refer to the second block of diagrams. There is shown the electromagnetic vertex diagram with one loop (generated by the photon "short-cutting between the electron legs") and two further diagrams with EM-vertex without loop, but each with one electron self-energy part on the in-coming electron leg respectively on the out-coming electron leg.
Is it true that (apart from the electron mass renormalization) taking these 3 diagrams into account, their respective divergent parts would cancel out mutually by virtue of the Ward identity ? One might argue that each single diagram is made finite at the end by the use of counter-terms, but my question refers to cancellation of their divergent parts without use of counter-terms.
Furthermore, could it be even concluded that the divergent part for any kind of complicated EM-vertex would cancel out against corresponding electron self-energy parts on the in-coming and out-coming leg of the graph for each order of $\alpha$ in perturbation theory by virtue of the Ward identity ? If it is not like this, which diagrams should be possibly added to achieve this or am I wrong with my understanding ? Thank you for any help.