What is Ao meets B1C ..........................

I think $K$ should be the intersection of $AA_1$ and $B_1C_1$. Let $A_1D$ be the diameter of the circumcircle of $\triangle A_1KB_1$. Let $M$ be the foot of perpendicular from $A_1$ to $CD$. Then $A_1KB_1DML$ is cyclic. Since $\angle DAC=90-\angle DAK=\angle KB_1A_1=2\angle HB_1A=2\angle CB_1D=2\angle LA_1D$, we have $\angle DA_1L=\angle CA_1L$, thus $DA_1=CA_1$. So the sum of the diameters is $A_1C+A_1B=BC$. Now since $\angle A_1LD=90=\angle CC_1B$ and $\angle A_1DL=\angle A_1B_1C=\angle CBC_1$, we have $\triangle A_1LD\sim\triangle CC_1B$ thus $\frac{A_1L}{CC_1}=\frac{A_1D}{BC}$. Similarly for the other side, add them up and using part (a), we get the result.