F4 Maths 309

AB = BC
∴ AB : BC = 1 : 1 △ABF ~ △ACE
∴ BF : CE = AB : AC = AB : (AB + BC) = 1 : (1 + 1) = 1 : 2 i.e. BF / CE = 1 / 2 ....... (1)
BC = 2CD
∴ CD : BC = 1 : 2 △CDG ~ △BDF
∴ CG : BF = CD : BD = CD : (CD + BC) = 1 : (1 + 2) = 1 : 3 i.e. CG / BF = 1 / 3 .......(2)
(1) * (2) :(BF / CE) (CG / BF) = (1 / 2) (1 / 3)CG / CE = 1 / 6CG : CE = 1 : 6

2011-12-26 12:29:11 補充：
Oh... the question is

CG : GE

= CG : (CE - CG)

= 1 : (6 - 1)

= 1 : 5 (Answer)

2011-12-26 12:51:54 補充：
Method 2 :

Let H be a point on EG such that FH // BC , then H is the mid point of CE.
So [[ CH = EH ]]

On the other hand , BC = FH and △FHG ~ △DCG .
∴ GH : CG = FH : DC = BC : DC = 2 : 1
So
CG : GE
= CG : (GH + EH)
= CG : (GH + CH)
= CG : (GH + CG+GH)
= 1 : (2 + 1+2)
= 1 : 5