中学数学竞赛题三角形ABC中的一点P,将P点分别以三边AB,AC和CB的中点为中心做反射,得到点Pc,Pb,Pa.（注意

连结AP,BP,CP,APa,BPb,CPc
显然四边形BPCPa是平行四边形
BP=CPa,CP=BPa,∠BCP=∠CBPa,∠CBP=∠BCPa
根据正弦定理
BPa/sin∠BAPa=APa/sin(∠ABC+∠CBPa)
CPa/sin∠CAPa=APa/sin(∠ACB+∠BCPa)
==>sin∠BAPa/sin∠CAPa
=CPsin(∠ABC+∠CBPa)/BPsin(∠ACB+∠BCPa)
=CPsin(∠ABC+∠BCP)/BPsin(∠ACB+∠CBP)同理
sin∠CBPb/sin∠ABPb
=APsin(∠ACB+∠CAP)/CPsin(∠BAC+∠ACP)
sin∠ACPc/sin∠BCPc
=BPsin(∠BAC+∠ABP)/APsin(∠ABC+∠BAP)
注意到∠ABC+∠BCP+∠BAC+∠ACP
=∠BCA+∠CAP+∠CBA+∠CBA
=∠BCA+∠CBP+∠BAC+∠ABP=π,代入得
(sin∠BAPa/sin∠CAPa)*(sin∠CBPb/sin∠ABPb)*(sin∠ACPc/sin∠BCPc)=1
根据角元ceva逆定理知APa,BPb,CPc共点