Let the decimal parts of $a/b$ and $c/d$ look like $A \overline{B}$ and $C \overline{D}$ respectively, for some strings $A, B, C, D$ with character set $\{0, 1, \dots, 9\}$. Further, we make the assumptions that $B, D$ have the smallest possible period call them $m, n$ and that $A, C$ have the shortest possible length for that period (i.e. the last character of $A$ ($C$) cannot be merged into the periodic part $\overline{B}$ ($\overline{D}$)). Consider the string $B^{|C| + |D|}$. It appears in $A\overline{B}$, thus it appears in $C\overline{D}$ as well. Since the suffix $B^{|D|}$ could not possibly intersect with $C$ due to $|B| \cdot |C| \geq |C|$, it follows that $B^{|D|}$ can be found in $D^{\infty}$. In particular, we obtain that for some cyclic shift $D'$ of $D$, we have $D'^{|B|} = B^{|D'|}$ forcing that $B, D$ are cyclic shifts of each other and that $|B| = |D|$ (they must have the same period). Replacing $C$ with a slightly larger string now if necessary, we assume $D = B$. Thus every finite substring from $A\overline{B}$ is found in $C\overline{B}$. Define $k = |C|$ and $n = |B|$ for ease of notation. We aim to show now that $A$ is a suffix of $C$. If $A$ is empty, we are already done, so assume this is not the case. We may assume $|C| > 0$ by adding a $D$ if necessary. Consider the strings $AB^{i}$ for positive integers $i$. Suppose the $i$th such string is found in $C\overline{B}$ from the $x_i$th position onward. Claim: We have $x_i > k$ for finitely many $i$. Proof: Assuming otherwise, suppose that $x_i > k$ for infinitely many $k$, so that $AB^{i}$ exists in $B^{\infty}$ for infinitely many $i$. Some $i \bmod n$ repeats infinitely often, fix such a residue $r$ and remark that due to periodicity it follows that $AB_1B_2\dots B_r B^{\infty}$ exists in $B^{\infty}$. But then it follows that $A$ completely extends into $B$, a contradiction. It follows that $x_i \in \{1, 2, \dots, k\}$ for large enough $i$. Therefore some $x_i$ repeats infinitely often, fix such an $x_i$ and observe that this then implies that $A$ agrees with $C_{x_i}C_{x_i+1}\dots C_k B_1 B_2 \dots B_{u}$ for some positive integer $u$. Since infinitely many $i$ are obtained by extending this very string, it follows that $u = 0$, otherwise $B$ would have a smaller period, or $A$ could be introduced into it. This shows that $A$ is a suffix of $C$. Let $\ell = |C|-|A|$ now and remark that $\frac{a}{b} - \frac{10^{\ell}c}{d}$ is an integer, and thus $b \mid d$. Repeating this argument from the other side, we obtain that $d \mid b$, forcing $b = d$ as required.