Answer: E

Small Hint:

Let $\mathbf d_n=\overrightarrow{A_nA_{n+1}}$ and derive a recurrence for these vectors

Big Hint:

Show that $\mathbf d_{n+4}=-\frac14\mathbf d_n,$ reducing the requested angle to one among the first few points

Solution:

Let $\mathbf d_n=\overrightarrow{A_nA_{n+1}}.$ The midpoint rule gives $\mathbf d_{n+3}=\frac12(\mathbf d_n+\mathbf d_{n+1})$ and also $\mathbf d_n+\mathbf d_{n+1}+\mathbf d_{n+2}=\frac12\mathbf d_n.$ Consequently, $$\begin{aligned} \mathbf d_{n+4} &=\frac12(\mathbf d_{n+1}+\mathbf d_{n+2})\\ &=-\frac14\mathbf d_n. \end{aligned}$$ Thus $\mathbf d_{43}$ and $\mathbf d_{44}$ are the same positive scalar multiple of $\mathbf d_3$ and $\mathbf d_4,$ respectively, so $\angle A_{44}A_{45}A_{43}=\angle A_4A_5A_3.$ Since $A_4$ and $A_5$ are the midpoints of $A_1A_2$ and $A_2A_3,$ $A_4A_5\parallel A_1A_3.$ The equilateral-triangle angles then give $\angle A_4A_5A_3=120^\circ.$

Therefore, the correct answer is E.