Triangle $ ABC$ has side lengths $ AB \equal{} 5$, $ BC \equal{} 6$, and $ AC \equal{} 7$. Two bugs start simultaneously from $ A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $ D$. What is $ BD$?

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$
1. Let's denote the points where the bugs meet as \( D \). Since the bugs start from \( A \) and crawl along the sides of the triangle \( ABC \) in opposite directions at the same speed, they will meet at a point \( D \) such that the distances they have crawled are equal.

2. Let the bug starting from \( A \) and crawling along \( AB \) reach point \( D \) after traveling a distance \( x \). Therefore, the bug starting from \( A \) and crawling along \( AC \) will have traveled a distance \( 7 - x \) to reach point \( D \).

3. Since the bugs meet at point \( D \), the total distance traveled by both bugs is equal to the perimeter of the triangle divided by 2. The perimeter of triangle \( ABC \) is:
   \[
   AB + BC + AC = 5 + 6 + 7 = 18
   \]
   Therefore, the total distance traveled by each bug is:
   \[
   \frac{18}{2} = 9
   \]

4. The bug traveling along \( AB \) reaches point \( D \) after traveling \( x \) units, and the bug traveling along \( AC \) reaches point \( D \) after traveling \( 7 - x \) units. Since they meet at point \( D \), the distance \( BD \) must be such that:
   \[
   x + (7 - x) + BD = 9
   \]
   Simplifying, we get:
   \[
   7 + BD = 9 \implies BD = 2
   \]

The final answer is \( \boxed{2} \).