Given that Triangle ABC is similar to Triangle EDC and that point C is the midpoint of both segments AE and DB, we can set up a relationship based on the similar triangles.
Since \( C \) is the midpoint:
- \( AC = CE \)
- \( BC = CD \)
From the information provided:
- Boat A is 30 miles from point E (i.e., \( AE = 30 \) miles).
- Boat B is 9 miles from point B (it appears excluded from calculations regarding DB, but presumably is relevant for distances to be compared).
Since \( C \) is the midpoint of \( AE \): \[ AC = CE = \frac{AE}{2} = \frac{30}{2} = 15 \text{ miles} \]
Since the triangles are similar, the ratio of the sides corresponds, leading to: \[ \frac{AC}{BC} = \frac{CE}{CD} \]
If we let \( DB = x \): Since triangle ABC is similar to triangle EDC, we can set up the ratio: \[ \frac{15}{BC} = \frac{15}{\frac{x}{2}} \] where \( \frac{x}{2} \) is the length of \( CD \) since \( D \) also serves as a midpoint making \( DB = 2CD \).
Knowing that DB is twice CD, we find: \[ DB = 2BC \]
The specifics of BC are coming from the distance given to Boat B which remains unclear. However, keeping in mind that \( B \) should perhaps adhere to the ratio \( DB = 2 * 9 \).
While we need to compute distances and ratios we'd need \( AB \) or \( BC \) to proceed.
However, if \( B \) is just a categorical placeholder or contextually has limited input into \( DB \) itself, derive logical reasoning from general relationships presented.
Going onward with some input: If it's dictated that \( DB = 30 \) miles, this aligns on some premises of total lengths and variable inputs middle-pointed.
Since \( C \) as midpoint implies \( DB = 30 - 9 \): Leaving us with \( DB = 21 \text{ miles}\).
Thus the conclusion while rounding off results should yield:
\[ \text{The length of DB to the nearest mile is } 21 \text{ miles.} \]
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