求第2题第3题咋么做,顺便看看图画对了吗?

画图正确,证明思路如下：（1）设垂足分别是D和D',若∠C=∠C’由△BCD≌△B'C'D'得CD=C'D',BC=B'C',∠CBD=∠C'B'D',又∵∠ABC=∠A'B'C',∴∠ABD=∠A'B'D',∴AD=A'D',∴AC=A'C',∴△ABC≌△A'B'C'（2）由△ABD≌△ACE得∠ADB=∠AEC,AD=AE,BD=CE,∠FBC=∠FCB得BF=CF,∴FD=FE,∴△ADF≌△AEF∴∠AFD=∠AFE,即FA平分∠EFD,∴点A到EF、DF的距离相等。