Let the foot of perpendicular from $B$ to $A_1C_1$ be $R$ .Let $\angle RBC_1 = x =\angle C_1A_1B$ Let the foot of perpendicular from $A$ on $C_1C_2$ and $C$ on $A_1A_2$ be $X$ and $Y$ be respectively.and Let $XA$ and $YC$ intersect at $J$ and let $\angle JBC =y$ we will select a mesurement such that $AB=sinC \Rightarrow BC =sinA$ and $CA =sinB$ It is not hard to calculate $tan x=\frac{BC_1}{BA_1}=\frac{sinC+sinB-sinA}{sinA+sinB-sinA}=\frac{CosA-sinA+1}{SinA-CosA+1}$ plugging $cosA=\frac{1-tan^2\frac{A}{2}}{1+tan^2\frac{A}{2}}$ and $sinA=\frac{2tan\frac{A}{2}}{1+tan^2\frac{A}{2}}$ and simplifyilg gives $tanx=\frac{1-tan\frac{A}{2}}{tan\frac{A}{2}(1+tan\frac{A}{2})}=cot\frac{A}{2}.tan(45-\frac{A}{2})$ Now apply the law of sines to$ \Delta BJC$ We have $\frac{JC}{siny}=\frac{JB}{sin (90-\frac{A}{2})}$(It is easy to calculate $\angle BCJ$) and similarly for $\Delta BAJ ,,,\frac{AJ}{cosy}=\frac{BJ}{sin(45+\frac{A}{2})}$ Dividing the two equations and simplifying $tany=\frac{JC}{JA}.\frac{sin(90-\frac{A}{2})}{sin(45+\frac{A}{2})}$ But by law of sines to $\Delta ACJ ,, \frac{JC}{AJ}=\frac{sin(45-\frac{A}{2})}{sin\frac{A}{2}}$ substuting this and simplifying $tany=cot\frac{A}{2}.tan(45-\frac{A}{2})=tanx$ Done!