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