三角形ABC中,角A=60°,I为内心,FI//AC交AB于F,3BP=BC,证：∠BFP=1/2∠B(要纯几何方法和向

如图,在△ABC中,∠A=60°,O为内心,OD//AC交AB于D,点E在BC上,且满足CE=2BE,连接DE,求证：∠B=2∠BDE.易知：A=60°=B/2+C/2过A作射线交BC于F,满足∠BAF=B/2, ∠CAF=C/2.于是要证原命题只需证：DE//AF延长BI交AC于G,则BD/BA=BI/BG=> BD/BA=(a+c)/(a+c+b) ...(1)BF/FC=c*sin(B/2) / b*sin(C/2)=> BF=ac*sinB/2 / (b*sinC/2+c*sinB/2)=> BE/BF=(b*sinC/2+c*sinB/2) / 3c*sinB/2C/2=60°-B/2代入得：BE/BF=1/3 + b*sin(60°) / 3c*tgB/2 - b/6c ...(2)rp=S=1/2 * bc * sin(60°)=> r=bc*sin(60°) / (a+b+c)tgB/2=r/ ((c+a-b)/2)=2bc*sin(60°) / (c+a-b)(a+b+c)代入(2)得：BE/BF=1/3 - b/6c +(a^2+c^2-b^2+2ac)/(6c^2) ...(3)利用a^2=b^2+c^2-bc,可以验证(1)=(3).=> BD/BA=BE/BF=> DE//AF=> ∠BDE=∠BAF=B/2=> ∠B=2∠BDE