答案：
(1) $\because AC=BC$，D为边AB的中点，$\therefore CD\perp AB$。$\because EF\perp AB$，$\therefore EF// CD$
(2) $\because AC=BC$，D为边AB的中点，$\therefore \angle ACD=\angle BCD$。$\because EF// CD$，$\therefore \angle ACD=\angle EGC$，$\angle BCD=\angle E$，$\therefore \angle EGC=\angle E$，$\therefore CG=CE$，$\therefore \triangle CEG$是等腰三角形

答案：
(1) $\because \angle B+\angle 1+\angle BDA=180^\circ$，$\angle B+\angle C+\angle BAC=180^\circ$，$\therefore \angle 1=180^\circ-\angle B-\angle BDA$，$\angle C=180^\circ-\angle B-\angle BAC$。$\because \angle BDA=\angle BAC$，$\therefore 180^\circ-\angle B-\angle BDA=180^\circ-\angle B-\angle BAC$，$\therefore \angle 1=\angle C$
(2) $\because BE$平分$\angle ABC$，$\therefore \angle ABE=\angle CBE=\frac{1}{2}\angle ABC$。$\because \angle BFD=\angle 1+\angle ABE$，$\angle AEF=\angle C+\angle CBE$，$\therefore \angle BFD=\angle AEF$。$\because \angle BFD=\angle AFE$，$\therefore \angle AEF=\angle AFE$，$\therefore AE=AF$

答案：过点O作$OD\perp AB$，垂足为D，过点O作$OE\perp AC$，垂足为E。$\because OD\perp AB$，$OE\perp AC$，AO平分$\angle BAC$，$\therefore OD=OE$。$\because \angle 1=\angle 2$，$\therefore OB=OC$。在$Rt\triangle BDO$和$Rt\triangle CEO$中，$\because Rt\triangle DOB\cong Rt\triangle EOC(HL)$，$\therefore \angle DBO=\angle ECO$，$\therefore \angle ABC=\angle ACB$，$\therefore AB=AC$