Question

如图,在△ABC中,AD,CE交于点F,BD:DC=1:2,F为AD的中点,求CF:FE的值

Answer 1

作FM平行于AB,交BC于M,则DM:MB=DF:FA=1:1,得DM=BM=(1/2)BD;

又BD:DC=1:2,则CD=2BD,故:CM:BM=(CD+DM):BM=(2BD+0.5BD):0.5BD=5:1.

所以,CF:FE=CM:BM=5:1.