Question

If $\frac{x}{l(mb+nc-la)}=\frac{y}{m(nc+la-mb)}=\frac{z}{n(la+mb-nc)},$ prove that $\frac{l}{x(by+cz-ax)}=\frac{m}{y(cz+ax-by)}=\frac{n}{z(ax+by-cz)}.$

Answer 1

We have $\frac{\frac{x}{l}}{mb+nc-la}=\frac{\frac{y}{m}}{nc+la-mb}=\frac{\frac{z}{n}}{la+mb-nc}$

And $\frac{\frac{y}{m}+\frac{z}{n}}{2la}=$ two similar expressions ;

$\therefore \frac{ny+mz}{a}=\frac{lz+nx}{b}=\frac{mx+ly}{c}.$

Multiply the first of these fractions above and below by $x,$ the second by $y,$ and the third by $z$; then

$\frac{nxy+mxz}{ax}=\frac{lyz+nxy}{by}=\frac{mxz+lyz}{cz}$

$=\frac{2lyz}{by+cz-ax}$

= two similar expressions ;

$\therefore \frac{l}{x(by+cz-ax)}=\frac{m}{y(cz+ax-by)}=\frac{n}{z(ax+by-cz)}.$