Re: [國中]比例線段與相似形
由 lskuo 於 星期二 十月 09, 2018 3:26 pm
3. 假設 BD:CD = a:1, AF:DF = b:1, 則
CDF:BDF = ACF:ABF = CD:BD = 1:a
ABF:BDF = AF:DF = b:1
所以 CDF:ABF = 1:ab
BCF:ABF = (CDF+BDF):ABF = (1+a):ab = CE:AE
AEF:ACF = AE:AC = ab:(1+a+ab)
EF:BF = AEF:ABF = ab/(1+a+ab)(ACF):ABF = b:(1+a+ab)
此題 a=3, and b=2,
BF:EF = (1+3+6):2 = 5:1
4. 假設三角形面積比 CFG:BDE:ADG = 1:a:b
作過A垂直BC的直線,分別交DG與BC於H,I 兩點。則
CF:BE:AH = 1:a:b
假設線段EF:CF = x:1, EI:CF = y:1, 則 FI:CF = (x-y):1
BE:BI = DE:AI = FG:AI = CF:CI
所以
a:(a+y) = x:(b+x) = 1:(1+x-y)
a:y = x:b = 1:(x-y)
xy = ab, x(x-y) = b, x^2 = b(a+1)
DEFG:CFG = EF^2:CF(EF)/2 = 2x:1
此例 CFG=2, BDE=3, ADG=4, 所以 a=3/2, b=2,
x^2 = 5,
所以 EDFG = (2x)(CFG) = 4x = 4(5)^(0.5)