![](http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/03/31be7c2376a6bd667d32b929482ab688.png)

- $\because \angle BAC=100^{\circ},BD=1$
  $\therefore$这是一个定线定角问题，可以使用辅助圆试试，辅助圆一般是这个角和这条边组成三角形的外接圆！

- 做$\triangle ABD$的外接圆,设交$BC$于$E$点

- 性质：圆的内接四边形对角互补
  $\therefore \angle BED=180^{\circ}-100^{\circ}=80^{\circ}$

  $\therefore \angle BDE=180^{\circ}-20^{\circ}-80^{\circ}=80^{\circ}$

  $\therefore BD=BE=1$

  双$\because \angle BED=\angle BCD+\angle CDE \Rightarrow 80^{\circ}=40^{\circ}+\angle EDC$
  $\therefore \angle EDC=40^{\circ}$

  $\therefore ED=EC=\sqrt{3}-1$

- 性质：圆周角相等对应的弦相等
  $\therefore AD=DE$

  $\therefore AD=\sqrt{3}-1$