KöMaL 2012 B4465

The line segment $A_0 A_{10}$ is divided into 10 equal parts. The dividing points are $A_{1}, A_{2}, \ldots, A_{9}$ in this order. The third vertex of the regular triangle drawn over the line segment $A_8 A_{10}$ is $B$. Show that $\angle BA_0 A_{10}$+$\angle BA_2 A_{10}$+$\angle BA_3 A_{10}$+$\angle BA_4 A_{10} =60^o$.