Question

相似三角形一道D,E分别在BC,AC上,AD,BE相交于点F,且CD=2BD,CE=3AE 则BF/EF=?

Answer 1

过D作DG‖BE;

则 EG/CE = BD/BC =1/3;

CE=3EG;

又∵CE=3AE ,

∴ EG = AE ;

即 AG=2AE;

∴ EF/DG = 1/2;

∵DG‖BE.

∴ DG/BE = CG/CE = (CE-EG)/CE = 2/3;

∴ BE/EF =(DG/BE)·(EF/DG) =3;

→ BF/EF = (BE-EF)/EF = (BE/EF)-1

=2