We assume that $C_{1}B_{1} \cap BC=K$ , where $C \in$ segment $BK$. We have two theorems: Cheva's and Menelaus theorem. By Cheva's theorem: $\frac{BA_{1}}{A_{1}C} \cdot \frac{CB_{1}}{B_{1}A}\cdot \frac{AC_{1}}{C_{1}B}=1$. By Menelaus's theorem: $\frac{KC}{KB} \cdot \frac{BC_{1}}{C_{1}A}\cdot \frac{AB_{1}}{B_{1}C}=1$. Hence, points $K,C,A_{1},B$ are harmonic and $\angle A_{1}DK=90^\circ$. So, $DA_{1}$ is bisector of angle $BDC$. Hence, implies $\frac{CD}{BD}=\frac{B_1C}{BC_1} \cdot \frac{C_1A}{AB_1}.$. Done

Any other way to do without harmonic??

@above then just extend $C_1$ and $B$ to $A_2$ and then use menelaus ( the ratio then interchanges wih ca2/ba2)

Let $T=B_1C_1\cap BC$, possibly at infinity, so that $(T,A_1;B,C)=-1$. As $C_1D\perp B_1C_1$, we conclude that $A_1D$ bisects $\angle BDC$. By the angle bisector theorem, we know that $\frac{CA_1}{BA_1}=\frac{CD}{BD}$. By Ceva, \[\frac{BA_{1}}{A_{1}C} \cdot \frac{B_{1}C}{B_{1}A}\cdot \frac{C_{1}A}{BC_{1}}=1 \implies \frac{B_1C}{BC_1} \cdot \frac{C_1A}{AB_1}=\frac{CA_1}{BA_1}=\frac{CD}{BD},\]as desired.