Vì : \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) tồn tại => a,b,c khác 0
Ta có : \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(\Rightarrow\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)
\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)
Áp dụng t/c dãy tỉ số bằng = nhau ta có :
\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)
\(=\dfrac{\left(abz-abz\right)+\left(bcx-bcx\right)+\left(acy-acy\right)}{a^2+b^2+c^2}=\dfrac{0}{a^2+b^2+c^2}=0\)
Do đó : \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}=0\)
\(\Rightarrow\left\{{}\begin{matrix}bz-cy=0\\cx-az=0\\ay-bx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{y}{b}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{y}{b}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
