Prove: Connect $B_{2} B_{3}$ intersecting $A_{1} B_{1}$ at $C_{1}$, draw the perpendiculars $B_{2} B_{2}^{\prime}, B_{3} B_{3}^{\prime}$ to $A_{2} A_{3}$. Then
$$\begin{array}{l} \frac{B_{2} C_{1}}{C_{1} B_{3}}=\frac{\mu_{3}}{\mu_{2}} \\ =\frac{\lambda_{1} /\left(\lambda_{3}+\lambda_{1}\right)}{\lambda_{1} /\left(\lambda_{1}+\lambda_{2}\right)} \end{array}$$
(by (1)
$$\begin{array}{l} =\frac{S_{\triangle B_{2} A_{2} A_{3}} / S_{\triangle A_{1} A_{2} A_{3}}}{S_{\triangle B_{3} A_{2} A_{3}} / S_{\triangle A_{1} A_{2} A_{3}}}=\frac{S_{\triangle B_{2} A_{2} A_{3}}}{S_{\triangle B_{3} A_{2} A_{3}}} \\ =\frac{B_{2} B_{2}^{\prime}}{B_{3} B_{3}^{\prime}} . \\ \text { Also } \frac{B_{2} C_{1}}{C_{1} B_{3}}=\frac{B_{2}^{\prime} B_{1}}{B_{1} B_{3}^{\prime}}, \\ \therefore \frac{B_{2}^{\prime} B_{1}}{B_{1} B_{3}^{\prime}}=\frac{B_{2} B_{2}^{\prime}}{B_{3} B_{3}^{\prime}} \\ \Rightarrow R_{1} \triangle B_{2} B_{2}^{\prime} B_{1} \sim \mathrm{R}_{1} \triangle B_{3} B_{3}^{\prime} B_{1} . \end{array}$$

Therefore,
$$\angle B_{2} B_{1} B_{2}^{\prime}=\angle B_{3} B_{1} B_{3}^{\prime} .$$

Furthermore, $\angle A_{1} B_{1} B_{2}=\angle A_{1} B_{1} B_{3}$.