比例知识
前言
注意引入非零比例因子的技巧的运用;
比例性质
- 合比定理
如果\(\cfrac{a}{b}=\cfrac{c}{d}\),那么\(\cfrac{a+b}{b}=\cfrac{c+d}{d}\),其中\(b,d\neq 0\);
证法1:由题目可知,\(\cfrac{a}{b}+1=\cfrac{c}{d}+1\),整理得到\(\cfrac{a+b}{b}=\cfrac{c+d}{d}\),
证法2:令\(\cfrac{a}{b}=\cfrac{c}{d}=k\),则\(a=bk\),\(c=dk\),代入得到
\(\cfrac{a+b}{b}=\cfrac{bk+b}{b}=k+1\),\(\cfrac{c+d}{d}=\cfrac{dk+d}{d}=k+1\),
故\(\cfrac{a+b}{b}=\cfrac{c+d}{d}\);
- 分比定理
如果\(\cfrac{a}{b}=\cfrac{c}{d}\),那么\(\cfrac{a-b}{b}=\cfrac{c-d}{d}\),其中\(b,d\neq 0\);
证法1:由题目可知,\(\cfrac{a}{b}-1=\cfrac{c}{d}-1\),整理得到\(\cfrac{a-b}{b}=\cfrac{c-d}{d}\),
证法2:同上;
- 合分比定理
如果\(\cfrac{a}{b}=\cfrac{c}{d}\),那么\(\cfrac{a+b}{a-b}=\cfrac{c+d}{c-d}\),其中\(b,d,a-b,c-d\neq 0\);
证明:令\(\cfrac{a}{b}=\cfrac{c}{d}=k\),则\(a=bk\),\(c=dk\),
则\(\cfrac{a+b}{a-b}=\cfrac{bk+b}{bk-b}=\cfrac{k+1}{k-1}\),\(\cfrac{c+d}{c-d}=\cfrac{dk+d}{dk-d}=\cfrac{k+1}{k-1}\),
故\(\cfrac{a+b}{a-b}=\cfrac{c+d}{c-d}\),
- 更比定理
如果\(\cfrac{a}{b}=\cfrac{c}{d}\),那么\(\cfrac{a}{c}=\cfrac{b}{d}\),其中\(a,b,c,d\neq 0\);
证明:令\(\cfrac{a}{b}=\cfrac{c}{d}=k\),则\(a=bk\),\(c=dk\),代入得到
\(\cfrac{a}{c}=\cfrac{bk}{dk}=\cfrac{b}{d}\);即\(\cfrac{a}{c}=\cfrac{b}{d}\);
应用举例
证明:引入非零比例因子,如\(\cfrac{a}{sinA}=\cfrac{b}{sinB}=\cfrac{c}{sinC}=2R\),
则\(a=2RsinA\),\(b=2RsinB\),\(c=2RsinC\),代入上式右端,得到
\(\cfrac{a+b-c}{sinA+sinB-sinC}=\cfrac{2R(sinA+sinB-sinC)}{sinA+sinB-sinC}=2R=\cfrac{a}{sinA}\);
故在\(\triangle ABC\)中,\(\cfrac{a}{sinA}=\cfrac{a+b-c}{sinA+sinB-sinC}\)成立;
同理,\(\cfrac{a}{sinA}=\cfrac{a+b+c}{sinA+sinB+sinC}\);\(\cfrac{a}{sinA}=\cfrac{a+b}{sinA+sinB}\);
分析:设\(2sinA=\sqrt{3}sinB=3sinC=k\),
则\(sinA=\cfrac{k}{2}\),\(sinB=\cfrac{k}{\sqrt{3}}\),\(sinC=\cfrac{k}{3}\),
则有\(a:b:c=sinA:sinB:sinC\),即\(a:b:c=\cfrac{k}{2}:\cfrac{k}{\sqrt{3}}:\cfrac{k}{3}=3:2\sqrt{3}:2\)
由此再设得到\(a=3m\),\(b=2\sqrt{3}m\),\(a=2m(m>0)\)(引入非零比例因子的好处),
由余弦定理可知,\(cosB=\cfrac{a^2+c^2-b^2}{2ac}=\cfrac{9m^2+4m^2-12m^2}{2\cdot 3m\cdot 2m}=\cfrac{1}{12}\)。
反思:1、灵活运用比例的性质,会大大简化运算;2、非零比例因子的引入,也要注意学习运用。
如图所示,三条直线\(l_1//l_2//l_3\),三条平行线与直线\(m\)分别相交于点\(A\)、\(B\)、\(C\),与直线\(n\)分别相交于点\(D\)、\(E\)、\(F\),连结\(AE\)、\(BD\)、\(BF\)、\(CE\),
根据平行线性质[等高]可得[利用等面积法],\(S_{\triangle ABE}=S_{\triangle DBE}\),\(S_{\triangle BEC}=S_{\triangle BEF}\),
\(\cfrac{S_{\triangle ABE}}{S_{\triangle DBE}}=\cfrac{S_{\triangle BEC}}{S_{\triangle BEF}}\),即\(\cfrac{S_{\triangle ABE}}{S_{\triangle BEC}}=\cfrac{S_{\triangle DBE}}{S_{\triangle BEF}}\),
根据等高三角形[\(\triangle ABE\)和\(\triangle BCE\)从顶点\(E\)所作的高线相同]的面积比等于底边的比,可得
由更比性质、等比性质可得,\(\cfrac{AB}{DE}=\cfrac{BC}{EF}=\cfrac{AB+BC}{DE+EF}=\cfrac{AC}{DF}\).
\(\cfrac{L}{L+M+N}\bar{x}\)\(+\)\(\cfrac{M}{L+M+N}\bar{y}\)\(+\)\(\cfrac{N}{L+M+N}\bar{z}\)\(=\)\(\cfrac{l}{l+m+n}\bar{x}\)\(+\)\(\cfrac{m}{l+m+n}\bar{y}\)\(+\)\(\cfrac{n}{l+m+n}\bar{z}\)
证明:由于 \(\cfrac{l}{L}\)\(=\)\(\cfrac{m}{M}\)\(=\)\(\cfrac{n}{N}\)\(=\)\(\cfrac{l+m+n}{L+M+N}\),
故 \(\cfrac{l}{l+m+n}\)\(=\)\(\cfrac{L}{L+M+N}\),\(\cfrac{m}{l+m+n}\)\(=\)\(\cfrac{M}{L+M+N}\),\(\cfrac{n}{l+m+n}\)\(=\)\(\cfrac{N}{L+M+N}\),
故有 \(\cfrac{L}{L+M+N}\bar{x}\)\(+\)\(\cfrac{M}{L+M+N}\bar{y}\)\(+\)\(\cfrac{N}{L+M+N}\bar{z}\)\(=\)\(\cfrac{l}{l+m+n}\bar{x}\)\(+\)\(\cfrac{m}{l+m+n}\bar{y}\)\(+\)\(\cfrac{n}{l+m+n}\bar{z}\)