\(x^2:(by+cz)=y^2:(cz+ax)=z^2:(ax+by)=1\) হলে, দেখাই যে, \(\cfrac{a}{a+x}+\cfrac{b}{b+y}+\cfrac{c}{c+z}=1\)
\(x^2:(by+cz)=y^2:(cz+ax)=z^2:(ax+by)=1 \)
বা, \(\cfrac{x^2}{(by+cz)}=\cfrac{y^2}{(cz+ax)}=\cfrac{z^2}{(ax+by)}=1\)
\(∴x^2=by+cz,y^2=cz+ax\) এবং
\( z^2=ax+by \)
এখন, \(\cfrac{a}{(a+x)}+\cfrac{b}{(b+y)}+\cfrac{c}{(c+z)}\)
\(=\cfrac{ax}{x(a+x)} +\cfrac{by}{y(b+y)} +\cfrac{cz}{z(c+z)} \)
\(=\cfrac{ax}{(ax+x^2 )}+\cfrac{by}{(by+y^2 )}+\cfrac{cz}{(cz+z^2 )}\)
\(=\cfrac{ax}{(ax+by+cz)}+\cfrac{by}{(by+cz+ax)}+\cfrac{cz}{(cz+ax+by)}\)
\(=\cfrac{(ax+by+cz)}{(ax+by+cz)}\)
\(=1\) (প্রমানিত)