题目如图,在Rt△ABCRt\triangle ABCRt△ABC中,∠ACB=90∘\angle ACB=90^{\circ}∠ACB=90∘,BCBCBC是⊙O\odot O⊙O的直径,⊙O\odot O⊙O与边ABABAB交于点DDD,EEE为BDBDBD的中点,连接CECECE,与ABABAB交于点FFF.(1)(1)(1)若∠A=4∠B\angle A=4\angle B∠A=4∠B,求∠ECB\angle ECB∠ECB的大小;(2)(2)(2)求证:AC=AFAC=AFAC=AF;(3)(3)(3)若BC=6BC=6BC=6,EFFC=12\frac{EF}{FC}=\frac{1}{2}FCEF=21,求△AFC\triangle AFC△AFC的面积.知识点:切线的判定章节:图形的性质 / 圆