II. (15 points) Given that the roots of the quadratic equation $a x^{2}+b x+c=0$ are 2013 times the roots of the quadratic equation $c x^{2}+d x+a=0$. Prove: $b^{2}=d^{2}$.

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Given that $a \neq 0, c \neq 0$.
Let $x_{1}, x_{2}$ be the two roots of $c x^{2}+d x+a=0$. Then $2013 x_{1}, 2013 x_{2}$ are both roots of the quadratic equation $a x^{2}+b x+c=0$.
By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-\frac{d}{c}, x_{1} x_{2}=\frac{a}{c} ; \\
2013 x_{1}+2013 x_{2}=-\frac{b}{a}, \\
2013 x_{1} \cdot 2013 x_{2}=\frac{c}{a} . \\
\text { Therefore, } \frac{2013 d}{c}=\frac{b}{a}, \frac{2013^{2} a}{c}=\frac{c}{a} .
\end{array}
$$

From the above two equations, we get
$2013^{2} \cdot \frac{d^{2}}{b^{2}}=\frac{c^{2}}{a^{2}}, 2013^{2}=\frac{c^{2}}{a^{2}}$.
Thus, $b^{2}=d^{2}$.