又是几何题
给一个麻烦的证法:角DCE+角DEC=角AFS+角DFS,而sinDCE/sinDEC=DE/CD=AB/CD=BS/SC=sinAFS/sinDFS。得证 过E作EX//AC交CD于X,则角DXE=FBS.
AD/BC=AF/CF
==>(AD*CD+AB*BC)/BC=(CF*AB+CD*AF)/CF
==>BF/BS=BF/BD*BD/BS=CF/AC*(AD*CD+AB*BC)/(AB*BC)=CF/AC+(CD*AF)/(AB*AC)=DX/EX+CD/EX=CX/EX
==>CEX和BFS相似
冏....:L 想到的同2楼,没觉得麻烦啊:L
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