(1)证明:连接A 交 C于点G,连接DG, 在正三棱柱ABC﹣![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105142-97360.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105142-20292.png) 中,四边形AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105143-77945.png) 是平行四边形, ∴AC=G , ∵AD=DB, ∴DG∥B![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105143-61661.png) ∵DG 平面 DC,B![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105144-28318.png) 平面 DC, ∴B ∥平面 DC. (2)解:过点D作DE⊥AC交AC于E,过点D作DF⊥ C交 C于F,连接EF. ∵平面ABC⊥面平AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105146-45079.png) ,DE 平面ABC,平面ABC∩平面AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105146-59234.png) =AC, ∴DE⊥平AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105146-49772.png) . ∴EF是DF在平面AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105147-42183.png) 内的射影. ∴EF⊥ C, ∴∠DFE是二面角D﹣ C﹣A的平面角, 在直角三角形ADC中, . 同理可求: . ∴ . ∴ . ∴![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105148-36793.png) |