51st Ukrainian National Mathematical Olympiad, 4th Round

In a triangle $ABC$ with $AC>BC>AB$. Points $D$ and $K$ are chosen on the sides $BC$ and $AC$ respectively so that $CD=AB$, $AK=BC$. $F$ and $L$ are the midpoints of the segments $BD$ and $KC$ respectively. $R$ and $S$ are the midpoints of the sides $AC$ and $AB$ respectively. The line segments $SL$ and $FR$ intersect at the point $O$, and it is known that $\angle SOF=55^{\circ }$. Find $\angle BAC$.