Same Dynkin diagrams $\Longrightarrow$ Isomorphic root systems. I am studying the book Introduction to Lie algebras. In page 122 there is something I don't understand and I am looking for some help.
In the beginning of that page the authors give the definition of what it means two root systems to be isomorphic:

Definition 1:
  Two root systems $R_1$ and $R_2$  in the real inner product spaces $E_1$ and $E_2$ respectively are isomorphic if there is a vector space isomorphism $\phi:E_1\to E_2$ such that:
  (a) $\phi(R_1)=R_2$,
  (b) $\color{brown}{ \text{ for any two roots }\alpha,\beta\in R_1, \ \langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle}$.

In the end of that page the authors prove a proposition saying that if the Dynkin diagrams of two root systems $R_1$ and $R_2$ are the same then the root systems are isomorphic.The proof begins as follows:
We may choose bases $B_1 = \{\alpha_1,\ldots , \alpha_\ell\}$ in $R_1$ and $B_2 = \{\alpha'_1,\ldots , \alpha'_\ell\}$ in $R_2$ so that for all $i, j$ one has
$\langle \alpha_i,\alpha_j\rangle =\langle \alpha'_i,\alpha'_j\rangle$.
Let $\phi: E \to E$ be the linear map which maps $\alpha_i$ to $\alpha'_i$. By definition, this is a vector space isomorphism satisfying $\color{brown}{\text{condition } 1(b)}$.
It is clear to me that for any two roots $\alpha,\beta\in B_1, \ \langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle$ but not clear at all that $\color{red}{\text{ for any two roots } \alpha,\beta\in R_1, \ \langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle}$. So can someone explain to me why this is true?
Note that here $\langle\alpha,\beta\rangle=\dfrac{2(\alpha,\beta)}{(\beta,\beta)}$ where $(,)$ is the inner product.
 A: Do you mean $2\frac{\langle\alpha,\beta\rangle}{\langle\beta,\beta\rangle}$ instead of $\langle\alpha,\beta\rangle$? Isomorphic root systems need not preserve the Killing form.
The proof that I'm familiar with works because the Weyl groups $W=\langle S_{\alpha_i}\rangle$ and $W'=\langle S_{\alpha_i'}\rangle$ are isomorphic, $\phi$ as you've defined is an isomorphism of $W$-modules, and $R_1=WB_1=\lbrace w(\alpha_i):w\in W\rbrace$ and $R_2=WB_2=\lbrace w(\alpha'_i):w\in W\rbrace$.
EDIT. Sorry, I hadn't seen that you'd clarified that you meant $2\frac{\langle\alpha,\beta\rangle}{\langle\beta,\beta\rangle}$.

EDIT 2. Perhaps this is a better way. If $\alpha\in B_1,\beta\in R_1$, then $\langle\alpha,\beta\rangle=\langle\phi(\alpha),\phi(\beta)\rangle$ by the linearity of $\phi$ and $\langle\cdot,\cdot\rangle$ in the second variable. If $\alpha\in R_1$, then $\alpha=w(\alpha_i)$, and
$$\phi(\alpha)=(\phi\circ w)(\alpha_i)=(\phi\circ\phi^{-1}\circ w'\circ\phi)(\alpha_i)=w'(\phi(\alpha_i))=w'(\alpha'_i)$$
and hence:
\begin{align*}
\langle\phi(\alpha),\phi(\beta)\rangle&=\langle w'(\alpha'_i),\phi(\beta)\rangle=\langle\alpha'_i,(w')^{-1}\phi(\beta)\rangle=\langle\alpha'_i,(\phi\circ w\circ\phi^{-1})^{-1}\phi(\beta)\rangle \\
&=\langle\alpha'_i,(\phi\circ w'\circ\phi^{-1}\circ\phi)(\beta)\rangle=\langle\phi^{-1}(\alpha'_i),w^{-1}(\beta)\rangle=\langle\alpha_i,w^{-1}(\beta)\rangle=\langle w(\alpha_i),\beta\rangle \\
&=\langle\alpha,\beta\rangle
\end{align*}
