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\(x^2:(by+cz)=y^2:(cz+ax)\)
\(=z^2:(ax+by)=1\)
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)