题目在Rt△ABCRt\triangle ABCRt△ABC中,∠C=90∘\angle C=90^{\circ}∠C=90∘,⊙O\odot O⊙O是△ABC\triangle ABC△ABC的内切圆,切点分别为DDD,EEE,FFF.(1)(1)(1)图111中三组相等的线段分别是CE=CFCE=CFCE=CF,AF=AF=AF=______,BD=BD=BD=______;若AC=3AC=3AC=3,BC=4BC=4BC=4,则⊙O\odot O⊙O半径长为______;(2)(2)(2)如图222,延长ACACAC到点MMM,使AM=ABAM=ABAM=AB,过点MMM作MN⊥ABMN\bot ABMN⊥AB于点NNN.求证:MNMNMN是⊙O\odot O⊙O的切线.知识点:切线的判定章节:图形的性质 / 圆