(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/20191022/20191022105154-27884.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105154-14804.png) 中,四边形AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105155-29024.png) 是平行四边形, ∴AC=G , ∵AD=DB, ∴DG∥B![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105156-53102.png) ∵DG 平面 DC,B![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105156-90583.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/20191022/20191022105158-11128.png) ,DE 平面ABC,平面ABC∩平面AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105159-62748.png) =AC, ∴DE⊥平AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105159-60925.png) . ∴EF是DF在平面AC![](http://img.shitiku.com.cn/uploads/allimg/20191022/20191022105200-33478.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: . ∴ . ∴ . ∴ . |