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

$r-q=\langle \beta,\alpha \rangle$
Note that the reflaction along $\alpha$, which  maps $\alpha\mapsto-\alpha$, $\beta\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha$ leaves the $\alpha$ string invariant, hence exchanges the ends of the string, i.e. 
$\beta-r\alpha\mapsto\beta+q\alpha$.
On the other hand, by linearity, $\beta-r\alpha\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha+r\alpha$ so that we conclude
$$r-q = 2\frac{\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}. $$
This shows that the second result $$ r-q=\langle\beta,\alpha\rangle$$ holds if and only if $\langle\alpha,\beta\rangle=0$ or $\langle\alpha,\alpha\rangle=2$.
Note that the reflaction along $\alpha$, which  maps $\alpha\mapsto-\alpha$, $\beta\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha$ leaves the $\alpha$ string invariant, hence exchanges the ends of the string, i.e. 
$\beta-r\alpha\mapsto\beta+q\alpha$.
On the other hand, by linearity, $\beta-r\alpha\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha+r\alpha$ so that we conclude
$$r-q = 2\frac{\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}. $$
This shows that the second result $$ r-q=\langle\beta,\alpha\rangle$$ holds if and only if $\langle\alpha,\beta\rangle=0$ or $\langle\alpha,\alpha\rangle=2$.