In the triangle $ ABC $, whose $ AC> BC> AB $, on the sides $ BC $ and $ AC $ chose the point $ D $ and $ K $, respectively, so that $ CD = AB $, $ AK = BC$. Points $ F $ and $ L - $ midpoints $ BD $ and $ CK $ respectively. Points $ R,S - $ midpoints of $ AC $ and $ AB $ respectively. Segments $ CL $ and $ FR $ intersect at point $ O $, with $ \angle SOF = 55^{\circ} $. Find $ \angle BAC $.
Using GeoGebra, I received a reply ($ \angle BAC= 70^{\circ} $), but I can not prove it strictly
$$\varphi =55^\circ ,AB=DC,AK=BC,AS=SB,BF=FD,KL=LC$$ $$BC=KL=a, CA=b,AB=CD=c$$
$$BF=FD=\frac{a-c}{2},KL=LC=\frac{b-a}{2}$$
by Menelaus's theorem: $\frac{AN}{NB}\cdot \frac{a-c}{a+c}=1 \Rightarrow NB=\frac{a-c}{2}=BF$
by Menelaus's theorem: $\frac{BM}{MC}\cdot \frac{b-a}{b+a}=1 \Rightarrow CM=\frac{b-a}{2}=CL$
$SR - $ middle line $\Rightarrow$
$$ \varphi = \alpha+\beta,\angle A=180^\circ-2\varphi=70^\circ$$