A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals:

[asy]size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F);
draw(D--G--H--A--cycle);
draw(Circle(O, abs(O-C)));
label("$A$", A, NW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SW);
label("$E$", E, SE);
label("$F$", F, SW);

label("4", (2,0.85), N);
label("3", D--E, S);
label("5", (6.5,0.85), N);
[/asy]
$\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$
1. First, observe that the two chords \( BC \) and \( EF \) are parallel. By drawing lines between \( B \) and \( E \) and also between \( C \) and \( F \), we form an isosceles trapezoid \( BCEF \).

2. Draw a perpendicular from point \( E \) to line \( AB \), and let the intersection point be \( G \). This forms a right triangle \( BEG \). Since \( DE = 3 \), and \( DE \) is parallel to \( AB \), the length \( AG \) is also \( 3 \). Therefore, \( BG = AB - AG = 4 - 3 = 1 \).

3. Similarly, draw a perpendicular from point \( F \) to line \( BC \), and let the intersection point be \( H \). This forms a right triangle \( CHF \). Since \( DE \) and \( EF \) are parallel, the length \( CH \) is also \( 1 \).

4. Now, we have two equal right triangles \( BEG \) and \( CHF \), since \( BE = CF \) and \( GE = HF \).

5. To find \( EF \), we sum the lengths \( BG \), \( BC \), and \( CH \):
   \[
   EF = BG + BC + CH = 1 + 5 + 1 = 7
   \]

The final answer is \(\boxed{7}\).