Let $E,E'$ be two euclidean vector spaces and $\Phi,\Phi'$ two root systems of $E$ and $E'$, respectively. Let $\varphi:E\to E'$ be an isomorphism of root systems. Applying the definition of root isomorphism, we easily check that $$\frac{(\varphi(\alpha),\varphi(\alpha))}{(\alpha,\alpha)}=\frac{(\varphi(\beta),\varphi(\beta))}{(\beta,\beta)},$$ for every pair of non orthogonal roots $\alpha,\beta\in\Phi$. I would like to show that this equality holds for all $\alpha,\beta\in\Phi$ when $\Phi$ is irreducible.
I am trying to show that there always exists a root $\gamma$ which is not proportional to $\alpha$ nor $\beta$ but I can't find the way to prove properly. I have also tried by using the fact that the Weyl's group natural action over an irreducible root system is irreducible.
I found a solution : given two roots $\alpha,\beta\in\Phi$, there exists a chain of roots $\gamma_0=\alpha,\gamma_1,\ldots,\gamma_r=\beta$ such that $(\gamma_i,\gamma_{i+1})\neq 0$, for all $1\leq i \leq r-1$. Indeed, write $\alpha\sim\beta$ when $\alpha,\beta$ are connected by such a chain. Fix $\alpha\in\Phi$ and put $\Phi_1=\{\beta\in\Phi\;:\; \alpha\sim\beta\}$ and $\Phi_2=\{\beta\in\Phi\;:\;\alpha\not\sim\beta\}$. Clearly, $\Phi=\Phi_1\sqcup\Phi_2$. Moreover, if $\gamma\in\Phi_1,\delta\in\Phi_2$ and $(\gamma,\delta)\neq 0$, we can add $\delta$ to the chain connecting $\alpha$ and $\gamma$, so that $\alpha\sim\delta$. This is a contradiction, hence $\Phi_1\perp\Phi_2$. As $\alpha\in\Phi_1$,$\Phi_1$ is nonempty, and the irreducibility of $\Phi$ then implies that $\Phi_2=\emptyset$. The result follows now directly.