△ABC,D在BC上,∠DAC=∠B,角平分线CE交AD于F,已知BD=1,DC=3,求CF:FE
问题描述：
△ABC,D在BC上,∠DAC=∠B,角平分线CE交AD于F,已知BD=1,DC=3,求CF:FE
答
,∠DAC=∠B ;,∠BCA=∠DCA =>△CBA~△CAD (aa) => CD:CA=3:CA=CA:CB=CA:4 =>CA=2√3AE:EB=AC:BC=2√3:4=√3 :2 ; DF:FA=CD:CA=3:2√3 =√3 :2梅涅劳斯 (CD/DB)*(BA/AE)*(EF/FC)=1=(3/1)*[(2+√3)/√3] *(EF/FC)=(2√...