比例性质
由上述性质以及其证明方法可以推广到任意的线性组合的比例性质。例如如下两条，分别从合分比性质和等比性质推广得到。
合分比性质的线性组合推论
已知
a
,
b
,
c
,
d
∈
C
{\displaystyle a,b,c,d\in \mathbb {C} }
，且有
b
≠
0
,
d
≠
0
{\displaystyle b\neq 0,d\neq 0}
，如果
a
b
=
c
d
{\displaystyle {\frac {a}{b}}={\frac {c}{d}}}
，则有
λ
a
+
μ
b
ξ
a
+
η
b
=
λ
c
+
μ
d
ξ
c
+
η
d
{\displaystyle {\frac {\lambda a+\mu b}{\xi a+\eta b}}={\frac {\lambda c+\mu d}{\xi c+\eta d}}}
。
证明：
∵
a
b
=
c
d
{\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}
令
k
=
a
b
=
c
d
{\displaystyle k={\frac {a}{b}}={\frac {c}{d}}}
，则
a
=
b
⋅
k
,
c
=
d
⋅
k
{\displaystyle a=b\cdot k,c=d\cdot k}
λ
a
+
μ
b
ξ
a
+
η
b
=
λ
b
k
+
μ
b
ξ
b
k
+
η
b
=
λ
k
+
μ
ξ
k
+
η
{\displaystyle {\frac {\lambda a+\mu b}{\xi a+\eta b}}={\frac {\lambda bk+\mu b}{\xi bk+\eta b}}={\frac {\lambda k+\mu }{\xi k+\eta }}}
λ
c
+
μ
d
ξ
c
+
η
d
=
λ
d
k
+
μ
d
ξ
d
k
+
η
d
=
λ
k
+
μ
ξ
k
+
η
{\displaystyle {\frac {\lambda c+\mu d}{\xi c+\eta d}}={\frac {\lambda dk+\mu d}{\xi dk+\eta d}}={\frac {\lambda k+\mu }{\xi k+\eta }}}
∴
λ
a
+
μ
b
ξ
a
+
η
b
=
λ
c
+
μ
d
ξ
c
+
η
d
{\displaystyle \therefore {\frac {\lambda a+\mu b}{\xi a+\eta b}}={\frac {\lambda c+\mu d}{\xi c+\eta d}}}
证毕
等比性质的线性组合推论一
已知
a
,
b
,
c
,
d
∈
C
{\displaystyle a,b,c,d\in \mathbb {C} }
，且有
b
≠
0
,
d
≠
0
{\displaystyle b\neq 0,d\neq 0}
，如果
a
b
=
c
d
{\displaystyle {\frac {a}{b}}={\frac {c}{d}}}
，则有
λ
a
+
μ
c
λ
b
+
μ
d
=
a
b
=
c
d
{\displaystyle {\frac {\lambda a+\mu c}{\lambda b+\mu d}}={\frac {a}{b}}={\frac {c}{d}}}
。
证明：
∵
a
b
=
c
d
{\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}
令
k
=
a
b
=
c
d
{\displaystyle k={\frac {a}{b}}={\frac {c}{d}}}
，则
a
=
b
⋅
k
,
c
=
d
⋅
k
{\displaystyle a=b\cdot k,c=d\cdot k}
λ
a
+
μ
c
λ
b
+
μ
d
=
λ
b
k
+
μ
d
k
λ
b
+
μ
d
=
k
{\displaystyle {\frac {\lambda a+\mu c}{\lambda b+\mu d}}={\frac {\lambda bk+\mu dk}{\lambda b+\mu d}}=k}
∴
λ
a
+
μ
c
λ
b
+
μ
d
=
a
b
=
c
d
{\displaystyle \therefore {\frac {\lambda a+\mu c}{\lambda b+\mu d}}={\frac {a}{b}}={\frac {c}{d}}}
证毕
等比性质的线性组合推论二
已知
a
i
,
b
i
∈
C
,
i
=
1
,
2
,
…
,
n
{\displaystyle a_{i},b_{i}\in \mathbb {C} ,i=1,2,\ldots ,n}
，且有
b
i
≠
0
{\displaystyle b_{i}\neq 0}
，如果
a
1
b
1
=
a
2
b
2
=
…
=
a
n
b
n
{\displaystyle {\frac {a_{1}}{b_{1}}}={\frac {a_{2}}{b_{2}}}=\ldots ={\frac {a_{n}}{b_{n}}}}
，则有
∑
i
n
λ
i
a
i
∑
i
n
λ
i
b
i
=
a
i
b
i
{\displaystyle {\frac {\sum _{i}^{n}{\lambda _{i}a_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {a_{i}}{b_{i}}}}
。
证明：
令
k
=
a
i
b
i
{\displaystyle k={\frac {a_{i}}{b_{i}}}}
，则
a
i
=
b
i
⋅
k
{\displaystyle a_{i}=b_{i}\cdot k}
∑
i
n
λ
i
a
i
∑
i
n
λ
i
b
i
=
∑
i
n
λ
i
k
b
i
∑
i
n
λ
i
b
i
=
k
∑
i
n
λ
i
b
i
∑
i
n
λ
i
b
i
=
k
{\displaystyle {\frac {\sum _{i}^{n}{\lambda _{i}a_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {\sum _{i}^{n}{\lambda _{i}kb_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {k\sum _{i}^{n}{\lambda _{i}b_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}=k}
∴
∑
i
n
λ
i
a
i
∑
i
n
λ
i
b
i
=
a
i
b
i
{\displaystyle \therefore {\frac {\sum _{i}^{n}{\lambda _{i}a_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {a_{i}}{b_{i}}}}
证毕