Let $A_1,B_1,C_1$ be the midpoints of the sides $BC,CA,AB$ respectively of a triangle $ABC$. Points $K$ on segment $C_1A_1$ and $L$ on segment $A_1B_1$ are taken such that $$\frac{C_1K}{KA_1}=\frac{BC+AC}{AC+AB}\enspace\enspace\text{and}\enspace\enspace\frac{A_1L}{LB_1}=\frac{AC+AB}{BC+AB}.$$If $BK$ and $CL$ meet at $S$, prove that $\angle C_1A_1S=\angle B_1A_1S$.