\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)
\(=\dfrac{abz-acy+cbx-abz+acy-cbx}{a^2+b^2+c^2}\)
\(=\dfrac{\left(abz-abz\right)+\left(acy-acy\right)+\left(cbx-cbx\right)}{a^2+b^2+c^2}\)
\(=\dfrac{0}{a^2+b^2+c^2}=0\)
Nên:
\(\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\)
Suy ra: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Hay \(x:y:z=a:b:c\left(đpcm\right)\)
Ta có:
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(\Leftrightarrow\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)
Áp dụng tính chất ta có:
\(\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)
\(=\dfrac{\left(abz-acy\right)+\left(cbx-abz\right)+\left(acy-cbx\right)}{a^2+b^2+c^2}\)
\(=\dfrac{0}{a^2+b^2+c^2}=0\)
\(\Rightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
hay \(x:y:z=a:b:c\left(đpcm\right)\)
