``````数学高手~~~SOS!
如图,D,E是在△ABC的边AB,AC上的交点,(AC>AB),且BD=CE, D,E的连线交BC的延长线于F.求证:AC*EF=AB*DF.A1 1 1 1 11 1 1 1 1D 1 1 1 1 EB 1 1 1 1 1 C 1 1 1 1 F. 大概是这样了.
参考答案:解:
从E。D。A点做CB的垂直线交CB分别是M,Q,N
得
CE/CA=ME/AQ
ND/AQ=BD/AB
EM/ND=EF/FD
左边相乘等于右边相成得:
CE*ND*EM/CA*AQ*ND=ME*BD*EF/AQ*AB*FD
解得:
AC*EF=AB*DF.