maths~~~~
2010-04-05 4:36 am
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回答 (2)
2010-04-05 5:44 am
✔ 最佳答案
CDE similar to GFECE : EG = DE : EF = 10 : 5 = 2 : 1
ABC similar to ADE
AE : AC = DE : BC = 10 : 6 = 5 : 3
let EG = x, AC = y
CE = 2x
CG = 2x - x = x
(y + 2x) : y = 5 : 3
3(y + 2x) = 5y
6x = 2y
3x = y
AC : CG : EG
= 3x : x : x
= 3 : 1 : 1
2010-04-05 5:47 am
AC/BC = AE/DE (property of similar triangle)
AC/6 = AE/10............(1)
AE = AC + CG + GE.........(2)
By mid- point theorem,
CG = GE ......................(3)
Sub. (2) and (3) into (1)
AC/6 = (AC + 2CG)/10
10AC = 6AC + 12 CG
4AC = 12CG
so AC : CG = 3 : 1
and AC : CG : GE = 3 : 1 : 1.
AC/6 = AE/10............(1)
AE = AC + CG + GE.........(2)
By mid- point theorem,
CG = GE ......................(3)
Sub. (2) and (3) into (1)
AC/6 = (AC + 2CG)/10
10AC = 6AC + 12 CG
4AC = 12CG
so AC : CG = 3 : 1
and AC : CG : GE = 3 : 1 : 1.
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