如图,△ABC中,AB>AC,∠A的平分线交外接圆于P,PE⊥AB,垂足为E,求证：AB-AC=2BE

Question

如图,△ABC中,AB>AC,∠A的平分线交外接圆于P,PE⊥AB,垂足为E,求证：AB-AC=2BE

Answer 1

证明：在AB上取一点F,使AF=AC,连接PF
△APF全等于△APC(SAS)
∠APF=∠APC
∠ABP=1/2弧ACP
∠PFB=∠PAB+∠APF
=∠PAC+∠APC
=1/2弧PC+1/2弧AC
=1/2弧ACP
=∠ABP
又PE⊥AB △PBF这等腰三角形,BE=EF
AB-AC=AB-AF=BF=2BE