（1）如图,连结CD,OC,则∠ADC =∠B = 60°．∵ AC⊥CD,CG⊥AD,∴∠ACG =∠ADC = 60°．由于∠ODC = 60°,OC = OD,∴△OCD为正三角形,得∠DCO = 60°．由OC⊥l,得∠ECD = 30°,∴∠ECG = 30° + 30° = 60°．进而∠A...