In $\triangle ABC$, if $\;$ $\dfrac{b + c}{11} = \dfrac{c + a}{12} = \dfrac{a + b}{13}$, then show that $\;$ $\dfrac{\cos A}{7} = \dfrac{\cos B}{19} = \dfrac{\cos C}{25}$
Let $\;$ $\dfrac{b + c}{11} = \dfrac{c + a}{12} = \dfrac{a + b}{13} = k \; (\text{say})$
Then, $b + c = 11 k$ $\;\;\; \cdots \; (1a)$, $\;$ $c + a = 12 k$ $\;\;\; \cdots \; (1b)$, $\;$ $a + b = 13 k$ $\;\;\; \cdots \; (1c)$
Subtracting equations $(1b)$ and $(1c)$ we have,
$b - c = k$ $\;\;\; \cdots (2)$
Adding equations $(1a)$ and $(2)$ we have,
$2 b = 12 k$ $\implies$ $b = 6 k$
Substituting the value of $b$ in equation $(1a)$ we have,
$c = 11 k - b = 11 k - 6 k = 5 k$
Substituting the value of $c$ in equation $(1b)$ we have,
$a = 12 k - c = 12 k - 5 k = 7 k$
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
\therefore \; \cos A & = \dfrac{\left(6 k\right)^2 + \left(5 k\right)^2 - \left(7 k\right)^2}{2 \times 6 k \times 5 k} \\\\
& = \dfrac{36 k^2 + 25 k^2 - 49 k^2}{60 k^2} \\\\
& = \dfrac{1}{5}
\end{aligned}$
$\begin{aligned}
\cos B & = \dfrac{\left(5 k\right)^2 + \left(7 k\right)^2 - \left(6 k\right)^2}{2 \times 5 k \times 7 k} \\\\
& = \dfrac{25 k^2 + 49 k^2 - 36 k^2}{70 k^2} \\\\
& = \dfrac{19}{35}
\end{aligned}$
$\begin{aligned}
\cos C & = \dfrac{\left(7 k\right)^2 + \left(6 k\right)^2 - \left(5 k\right)^2}{2 \times 7 k \times 6 k} \\\\
& = \dfrac{49 k^2 + 36 k^2 - 25 k^2}{84 k^2} \\\\
& = \dfrac{5}{7}
\end{aligned}$
Now,
$\begin{aligned}
\cos A : \cos B : \cos C & = \dfrac{1}{5} : \dfrac{19}{35} : \dfrac{5}{7} \\\\
& = \dfrac{7}{35} : \dfrac{19}{35} : \dfrac{25}{35}
\end{aligned}$
$\implies$ $\dfrac{\cos A}{7} = \dfrac{\cos B}{19} = \dfrac{\cos C}{25}$
Hence proved.