$\begin{array}{l}\quad \dfrac{1}{c} = \dfrac{1}{2}\left(\dfrac{1}{a} + \dfrac{1}{b}\right)\qquad (a;\,b;\,c\ne 0)\\ \to \dfrac{2}{c} =\dfrac{a+b}{ab}\\ \to 2ab = c(a + b)\\ \to ab - bc = ac - ab\\ \to b(a-c) = a(c-b)\\ \to \dfrac{a}{b} = \dfrac{a-c}{c-b} \end{array}$