已知四边形ABCD内接于圆O,AB与CD的延长线交于点E,AD与BC的延长线交于点F,EG、FG分别是角AEC和角AFC的角平分线.求证:EG⊥FG.

问题描述:

已知四边形ABCD内接于圆O,AB与CD的延长线交于点E,AD与BC的延长线交于点F,EG、FG分别是角AEC和角AFC的角平分线.求证:EG⊥FG.

DAE=DFC+ABFDCF=AED+CBECBE=ABF=CFE+AEF圆内接四边形对角互补,DAE+DCF=180所以:DFC+(CFE+AEF)+AED+(CFE+AEF)=180DFC/2+AED/2+(CFE+AEF)=90GFC+GED+(CE+AEF)=90GFE+GEF=90FGE=90