প্রদত্ত সমীকরণ: $$ \begin{aligned} \frac{1}{a+c} + \frac{1}{b+c} &= \frac{3}{a+b+c} \end{aligned} $$ $$ \begin{aligned} \frac{a+b+2c}{(a+c)(b+c)} &= \frac{3}{a+b+c} \end{aligned} $$ আড়াআড়ি গুণ করে পাই: $$ \begin{aligned} (a+b+2c)(a+b+c) &= 3(ab + ac + bc + c^2) \end{aligned} $$ $$ \begin{aligned} (a+b)^2 + 3c(a+b) + 2c^2 &= 3ab + 3ac + 3bc + 3c^2 \end{aligned} $$ $$ \begin{aligned} a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 &= 3ab + 3ac + 3bc + 3c^2 \end{aligned} $$ $$ \begin{aligned} a^2 + b^2 - c^2 &= ab \end{aligned} $$ কোসাইন সূত্র অনুযায়ী, $$ \begin{aligned} \cos C &= \frac{a^2 + b^2 - c^2}{2ab} = \frac{ab}{2ab} = \frac{1}{2} \end{aligned} $$ অতএব, $C = 60^\circ$।