We are given that $\triangle ABC \cong \triangle EDC$. This means that the corresponding sides and angles of these two triangles are congruent. We are also given that $C$ is the midpoint of $\overline{AE}$ and $\overline{DB}$. This means $AC = CE$ and $DC = CB$. We are told that the distance between Boat A and Boat B is 5 miles. This can be represented as $AB = 5$. The distance between Boat A and point E is 30 miles, so $AE = 30$.
Since C is the midpoint of AE, we have $AC = CE = \frac{1}{2} AE$. We are given that $AE = 30$, so $AC = \frac{1}{2} \times 30 = 15$.
Since C is the midpoint of DB, we know that $DC = CB$. Since the triangles ABC and EDC are congruent, their corresponding sides are equal. Therefore, $AC = CE$ and $BC = CD$. In triangle ABC, $AC = 15$. Since $AC = CE$, it follows that $CE = 15$. This is consistent with $AE = AC + CE = 15 + 15 = 30$. We are given $AB = 5$. We also know that the triangles are congruent, so $AB = DE = 5$.
Since the triangles are congruent, we have $BC = CD$. Therefore, $DB = DC + CB = 2 \times BC$.
Triangles ABC and CDE are congruent. This implies that $AB = DE = 5$ and $AC = CE = 15$. Also, $BC = CD$. Since $AB=5$, we can infer something about $BC$. However, we don't have any angles, so using the Pythagorean theorem is not possible. But we know that $DB = BC + CD$, and since $C$ is the midpoint, $BC = CD$. Therefore $DB = 2 \times BC = 2 \times CD$. We know that the triangles are congruent. Since AC is 15, and CE is 15 (C being the midpoint) and $AE=30$. Similarly, $BC = CD$. Thus, $DB = 2 \times BC$. Since $\triangle ABC \cong \triangle EDC$, $BC = CD$. We are given that $AB = 5$ so $DE = 5$. Since $AC=15$ and triangle EDC is congruent to triangle ABC, so $CD$ must be much smaller than 15 to make $\triangle EDC$ fit into the picture.
Since $\triangle ABC \cong \triangle EDC$, we have $AB = DE = 5$ and $AC = CE = 15$. Also, $BC = CD$. Since C is the midpoint of DB, $DB = 2 * BC$. Since $\triangle ABC \cong \triangle EDC$, the corresponding sides are equal, so $BC = CD$. Since $C$ is the midpoint of $DB$, we have $DB = 2*BC$. However, we are given $AB=5$, so using similarity we see that the scaling of the bigger triangle is a factor of $\frac{AE}{AB} = \frac{30}{5} = 6$. Thus, $BC \cdot 6 = AC$ implies $BC = \frac{15}{6} = \frac{5}{2}$. So $DB = 2 BC = 2 \cdot \frac{5}{2} = 5$. But we are looking at $BC = CD$ so $BD = 2 \times BC$. Then we have $DB=2BC$.
Since we know $AC = 15$ and the scale factor is 6 between the two triangles, we must have $BC = \frac{15}{6} = 2.5$. Therefore $DB = 2(2.5) = 5$.
\\(\boxed{5 \text{ miles}}\\)
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