(1)证明: ∵2∠ACD=90°, ∴∠ACD=45° ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105040728-95791.png) ∵∠DOC=90°,且DO=CO, ∴△OCD为等腰直角三角形,∠OCD=45° ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105040728-32143.png) ∴∠ACO=∠ACD+∠DCO=45°+45°=90° ∴直线AC是⊙O的切线. ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105040728-26031.png) (2)解:连接BO, ∵∠ACB=75°,∠ACD=45°, ∴∠DCB=30°,∴∠DOB=60°, ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105040728-12851.png) ∵DO=BO, ∴△BDO为等边三角形, ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105040728-19787.png) ∴BD=OB=4. ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105040729-89390.png) (1)利用切线的判定定理求出∠ACO=∠ACD+∠DCO=45°+45°=90°,即可得出答案; (2)利用圆周角定理得出△BDO为等边三角形,即可得出答案. |