^2+\delta_{B_2{'}}^2+\delta_{A_2{'}}\cdot\delta_{B_2{'}})}, &&\text{(3.7)} \\ &m_{lga_x}=\sqrt{m_{lgb}^2+\frac{2}{3}m^{n2}\sum_l^{n-1}(\delta_{A_2{'}}^2+\delta_{B_2{'}}^2+\delta_{A_2{'}}\cdot\delta_{B_2{'}})+\delta_{C_2{'}}^2+\delta_{B_2{'}}^2\delta_{C_2{'}}^2\cdot\delta_{B_2{'}}^2}, &&\text{(3.8)} \end{align} </math>

підставляючи (3.7), (3.8) у (3.6), отримаємо:

${\displaystyle \sum _{l}^{n}(\delta _{A}^{2}+\delta _{B}^{2}+\delta _{A}^{2}\cdot \delta _{B}^{2})=\sum _{l}^{n-1}(\delta _{A}^{2}+\delta _{B}^{2}+\delta _{A}^{2}\cdot \delta _{B}^{2})+\delta _{C_{x}}^{2}+\delta _{B_{x}}^{2}+\delta _{C_{x}}\cdot \delta _{B_{x}}}$

Тобто

${\displaystyle \delta _{A_{x}}^{2}+\delta _{A_{x}}\delta _{B_{x}}+\delta _{B_{x}}^{2}=\delta _{C_{x}}^{2}+\delta _{C_{x}}+\delta _{B_{x}}+\delta _{B_{x}}^{2}}$,

звідки

${\displaystyle {\begin{aligned}\delta _{A_{x}}^{2}+\delta _{A_{x}}\delta _{B_{x}}=\delta _{C_{x}}^{2}+\delta _{C_{x}}^{2}+\delta _{B_{x}}&&{\text{(3.9)}}\\\end{aligned}}}$

Эамінюючи в цьому рівнянні величини δ їх значеннями з виразу:

${\displaystyle {\begin{aligned}\delta _{A_{x}}={\frac {M}{\rho ^{n}}}ctgA&&{\text{(3.10)}}\\\end{aligned}}}$

і поділивши ліву і праву частини рівності на ${\displaystyle {\frac {M}{\rho ^{n}}}}$, отримаємо:

${\displaystyle ctg^{2}A_{n}+ctgA_{n}ctgB_{n}=ctg^{2}C_{n}+ctgC_{n}ctgB_{n}}$.

Така рівність можлива, якшо:

${\displaystyle ctgA_{n}=ctgC_{n}}$

і тому

${\displaystyle {\begin{aligned}A_{n}=C_{n}&&{\text{(3.11)}}\end{aligned}}}$

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