In a rhombus, $ ABCD$, line segments are drawn within the rhombus, parallel to diagonal $ BD$, and terminated in the sides of the rhombus.  A graph is drawn showing the length of a segment as a function of its distance from vertex $ A$.  The graph is:

$ \textbf{(A)}\ \text{A straight line passing through the origin.} \\
\textbf{(B)}\ \text{A straight line cutting across the upper right quadrant.} \\
\textbf{(C)}\ \text{Two line segments forming an upright V.} \\
\textbf{(D)}\ \text{Two line segments forming an inverted V.} \\
\textbf{(E)}\ \text{None of these.}$
1. Let \( d \) denote the distance from vertex \( A \) measured along diagonal \( AC \), and let \( l(d) \) denote the length of the segment parallel to diagonal \( BD \) and \( d \) units from \( A \).

2. Consider the case when \( d \le \frac{AC}{2} \). By similar triangles, we have:
   \[
   \frac{\frac{1}{2} l}{d} = \frac{\frac{1}{2} BD}{\frac{1}{2} AC}
   \]
   Simplifying this, we get:
   \[
   \frac{l}{2d} = \frac{BD}{AC}
   \]
   Let \( k = \frac{BD}{AC} \). Then:
   \[
   \frac{l}{2d} = k \implies l = 2kd
   \]

3. Now consider the case when \( d \ge \frac{AC}{2} \). Again, by similar triangles, we have:
   \[
   \frac{\frac{1}{2} l}{AC - d} = \frac{\frac{1}{2} BD}{\frac{1}{2} AC}
   \]
   Simplifying this, we get:
   \[
   \frac{l}{2(AC - d)} = \frac{BD}{AC}
   \]
   Using \( k = \frac{BD}{AC} \), we get:
   \[
   \frac{l}{2(AC - d)} = k \implies l = 2k(AC - d)
   \]
   Expanding this, we get:
   \[
   l = 2kAC - 2kd
   \]

4. The graph of \( l \) as a function of \( d \) is linear in \( d \). For \( d < \frac{AC}{2} \), the slope is \( 2k \), which is positive. For \( d > \frac{AC}{2} \), the slope is \( -2k \), which is negative. Therefore, the graph forms an inverted V shape.

Conclusion:
\[
\boxed{\textbf{(D)}}
\]