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With usual notations, if in a triangle `A B C(b+c)/(11)=(c+a)/(12)=(a+b)/(13)` , then prove that: `(cosA)/7=(cosB)/(19)=(cosC)/(25)`

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Let `(b+c)/11 = (c+a)/12 = (a+b)/13 = k`
`=>b+c = 11k->(1)`
`=>c+a = 12k->(2)`
`=>a+b = 13k->(3)`
Solving (1),(2) and (3), we get,
`a = 7k, b = 6k, c = 5k`
Now we know the formula of `cos` is, `cosA = (b^2+c^2-a^2)/(2bc) = (36k^2+25k^2-49k^2)/(60k^2) = 1/5`
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