等腰△ABC中,AB=AC,AD⊥BC于D,CG‖AB,BG分别交AD、AC于E、F.

证法一：连接CE
1) AB=AC,AD⊥BC可知∠BAD=∠CAD,根据SAS可知△ABE≌△ACE,于是BE=CE,∠ABE=∠ACE
2) CG//AB可知∠CGE=∠ABE=∠FCE,又∠CEG=∠FEC（公共角）,于是△CGE相似于△FCE,从而EC^2=EF*EG
3) 因为BE=CE,所以BE^2=EF*EG
证法二：延长GC交AD的延长线于H,连接BH
1) AB=AC,AD⊥BC可知∠BAD=∠CAD、BD=CD,又DH⊥BC可知△BDH≌△CDH,从而∠BHD=∠CHD
2) CH//AB可知∠CHD=∠BAD,于是∠BHD=∠CHD=∠BAD=∠CAD,从而BH//AC
3) BH//AC得BE/EF=EH/AE,GH//AB得EG/BE=EH/AE,从而BE/EF=EG/BE,变形即得BE^2=EF*EG