Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$
Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is 
\[2d + d = 3d\]
and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we create two different right triangles: 
- One with base $\frac{38}{2}= 19$, height $d$, and hypotenuse $r$  ($\triangle RAO$ on the diagram)
- Another with base $\frac{34}{2} = 17$, height $3d$, and hypotenuse $r$  ($\triangle LBO$ on the diagram)
By the Pythagorean theorem, we can create the following system of equations:
\[19^2 + d^2 = r^2\]
\[17^2 + (2d + d)^2 = r^2\]
Solving, we find $d = 3$, so $2d = \boxed{\textbf{(B)}\ 6}$.
-Solution by Joeya, diagram by Jamess2022(burntTacos), and minor edits by lpieleanu.