Ta có : 

  \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cxy}{ax}=\frac{cxy-ayz}{by}\)

             \(=\frac{ayz-bxz}{cz}=\frac{0}{ax+by+cz}=0\)

\(\Leftrightarrow bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\)          \(\left(1\right)\)

     \(cx=az\Rightarrow\frac{x}{a}=\frac{z}{c}\)           \(\left(2\right)\)

     \(ay=bx\Rightarrow\frac{y}{b}=\frac{x}{a}\)           \(\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) hay \(x:y:z=a:b:c\)

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{abx-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)

\(=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

=>bz-cy=cx-az=ay-bx=0

=>\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\Rightarrow a:b:c=x:y:z\)(đpcm)

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

=>\(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

=>\(\hept{\begin{cases}\frac{bz-cy}{a}=0\\\frac{cx-az}{b}=0\\\frac{ay-bx}{c}=0\end{cases}\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Rightarrow}\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}\Rightarrow}\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{b}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}\)

hay x:y:z=a:b:c

ai xoạc nào

Ta có \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(\Rightarrow\frac{\left(bz-cy\right).x}{ax}=\frac{\left(cx-az\right)y}{by}=\frac{\left(ay-bx\right).z}{cz}\)

\(\Rightarrow\frac{bxz-cxy}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cxy+cxy-ayz+ayz-bxz}{ax+by+cz}\)

Suy ra:

        bz - cy = 0                        (1)

        cx - az = 0                        (2)

        ay - bx = 0                        (3)

Từ (1) ta có: \(bz=cy\Rightarrow\frac{y}{b}=\frac{z}{c}\left(I\right)\)

Từ (2) ta có: \(cx=az=\frac{z}{c}=\frac{x}{a}\left(II\right)\)

Từ (3) ta có: \(ay=bx=\frac{x}{a}=\frac{y}{b}\left(III\right)\)

Từ (I), (II), (III) => x: y: z = a: b: c

\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

\(=\dfrac{a\left(bz-cy\right)}{a.a}=\dfrac{b\left(cx-az\right)}{b.b}=\dfrac{c\left(ay-bx\right)}{c.c}\)

\(=\dfrac{abz-acy}{a^2}=\dfrac{bcx-baz}{b^2}=\dfrac{cay-cbx}{c^2}\)

\(=\dfrac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}\)

\(=\dfrac{0}{a^2+b^2+c^2}\)

\(=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bz-cy}{a}=0\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{cx-az}{b}=0\Rightarrow cx-az=0\Rightarrow cx=az\Rightarrow\dfrac{c}{z}=\dfrac{a}{x}\\\dfrac{ay-bx}{c}=0\Rightarrow ay-bx=0\Rightarrow ay=bx\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

\(\Rightarrow a:b:c=x:y:z\)