$\triangle ABC$ 中，$A=150^\circ$，$D_i$ $\left(i=1,2,\cdots,2020\right)$ 为选段 $BC$ 上的点，满足 $BD_1=D_1D_2=\cdots=D_{2019}D_{2020}=D_{2020}C$．设 $\angle BAD_1=\alpha_1$，$\angle D_1AD_2=\alpha_2$，$\cdots$，$D_{2019}AD_{2020}=\alpha_{2020}$，$\angle D_{2020}AC=\alpha_{2021}$．求 $\frac{\sin \alpha_1\sin \alpha_3\cdots\sin \alpha_{2021}}{\sin\alpha_2\sin\alpha_4\cdots\sin\alpha_{2020}}$ 的值．
【难度】
【出处】
2020年北大强基考试数学回忆
【标注】
【答案】
略
【解析】
由正弦定理容易得到 $\frac{\sin\alpha_1}{\sin\alpha_2}=\frac{AD_2}{AB}$，$\frac{sin\alpha_3}{\sin\alpha_4}=\frac{AD_4}{AD_2}$ $\cdots$ $\frac{\sin\alpha_{2019}}{\sin\alpha_{2020}}=\frac{AD_{2020}}{AD_{2018}}$
$\frac{\sin \alpha_1\sin \alpha_3\cdots\sin \alpha_{2021}}{\sin\alpha_2\sin\alpha_4\cdots\sin\alpha_{2020}}=\frac{AD_{2020}\sin\alpha_{2021}}{AB}=\frac{D_{2020}C\sin C}{AB}=\frac{1}{2021}\frac{BC\sin C}{AB}=\frac{\sin A}{2021}=\frac{1}{4042}$．
$\frac{\sin \alpha_1\sin \alpha_3\cdots\sin \alpha_{2021}}{\sin\alpha_2\sin\alpha_4\cdots\sin\alpha_{2020}}=\frac{AD_{2020}\sin\alpha_{2021}}{AB}=\frac{D_{2020}C\sin C}{AB}=\frac{1}{2021}\frac{BC\sin C}{AB}=\frac{\sin A}{2021}=\frac{1}{4042}$．
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