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I am trying to solve following exercise:

Let $\alpha,\beta\in\Phi$. Let the $\alpha$-string through $\beta$ be $\beta-r\alpha,\cdots, \beta+q\alpha$. Let the $\beta$-string through $\alpha$ be $\alpha-r'\beta$, $\cdots$, $\alpha+q'\beta$. Prove that $$\frac{q(r+1)}{(\beta,\beta)}=\frac{q'(r'+1)}{(\alpha,\alpha)}.$$

This has already appeared thrice in mathstackexchange (first comment below), and one answer was not understandable (second comment below).

According to other answer, it was suggested to consider possible root systems and verify this identity for each root system. I did the verification for each root system of tpe $A_1\times A_, A_2, B_2, G_2$.

Question: Can we prove the above identity without using classification of root systems?

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    http://math.stackexchange.com/questions/46513/on-the-root-systems?rq=1, then http://math.stackexchange.com/questions/561806/equation-on-root-systems-humphreys-exercise-9-10 and http://math.stackexchange.com/questions/319534/prove-that-fracqr1-beta-beta-fracqr1-alpha-alpha – 2017-01-20
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    Among three answer to earlier similar questions, one answer had perhaps made wrong use of symbols $(\alpha,\beta)$ and $\langle \alpha,\beta\rangle$ - see third link above; so I neglected it. Also it was too long back; I don't know if that can be revisited? – 2017-01-20

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http://www.math.ualberta.ca/~zhchang/Research/Notes_01_STLA.pdf

Here is an other solution to the problem, I don't think it uses the clasification theorem but I'm not sure since I don't know what they mean by $e_{\alpha}$.

Actually I just realized that in proposition 25.2 of Humphreys Lie algebra book appears this proof.

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    (It's on page 35/36.) This is a neat argument that avoids the classification, but it uses that every root system comes from a Lie algebra and properties of a Chevalley basis (whereof the $e_\alpha$ are members), which is not such an obvious result either. – 2017-11-17