(1)连接OC,如图①所示, ∵OC=OA, ∴∠BAC=∠OCA, ∵EF切⊙O于C, ∴OC⊥EF,又AD⊥EF, ∴OC∥AD, ∴∠OCA=∠DAC, ∴∠DAC=∠BAC; (2)∠BAG=∠DAC,理由如下: 连接BC,如图②所示, ∵AB为⊙O的直径, ∴∠BCA=90°, ∴∠B+∠BAC=90°, ∵AD⊥EF,∴∠ADG=90°, ∴∠AGD+∠GAD=90°, 又![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105120632-34815.png) | AC | =![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105120632-34815.png) | AC | ,∴∠B=∠AGD, ∴∠BAC=∠GAD, ∴∠BAG+∠GAC=∠GAC+∠DAC,即∠BAG=∠DAC.
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105120633-59578.png) |