Cho (bz - cy)/a = cx - az/b= ay - bx/c. Chứng minh rằng x/a =y/b =z/c

Ta có: \(\frac{{bz - cy}}{a} = \frac{{abz - acy}}{{{a^2}}}\);

\(\frac{{cx - az}}{b} = \frac{{bcx - baz}}{{{b^2}}}\); \(\frac{{ay - bx}}{c} = \frac{{cay - cbx}}{{{c^2}}}\).

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

Nên \(\frac{{bz - cy}}{a} = \frac{{ay - bx}}{c} = \frac{{abz - acy}}{{{a^2}}} = \frac{{bcx - baz}}{{{b^2}}} = \frac{{cay - cbx}}{{{c^2}}}\)

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

Do đó \[bz - cy = 0;\,\,ay - bx = 0\].

Khi đó, \(bz = cy\) nên \(\frac{b}{y} = \frac{c}{z}\) và \(ay = bx\) nên \(\frac{b}{y} = \frac{a}{x}\).

Do đó \(\frac{a}{x} = \frac{b}{y} = \frac{c}{z}\) (đpcm).