In rectangle $ ABCD$, $ AB\equal{}100$. Let $ E$ be the midpoint of $ \overline{AD}$. Given that line $ AC$ and line $ BE$ are perpendicular, find the greatest integer less than $ AD$.
1. **Place the rectangle in the coordinate plane:**
   - Let \( A = (0, b) \), \( B = (100, b) \), \( C = (100, 0) \), and \( D = (0, 0) \).
   - Since \( E \) is the midpoint of \( \overline{AD} \), the coordinates of \( E \) are \( (0, \frac{b}{2}) \).

2. **Calculate the slopes of lines \( AC \) and \( BE \):**
   - The slope of \( AC \) is:
     \[
     \text{slope of } AC = \frac{0 - b}{100 - 0} = -\frac{b}{100}
     \]
   - The slope of \( BE \) is:
     \[
     \text{slope of } BE = \frac{\frac{b}{2} - b}{0 - 100} = \frac{-\frac{b}{2}}{-100} = \frac{b}{200}
     \]

3. **Use the condition that \( AC \) and \( BE \) are perpendicular:**
   - Two lines are perpendicular if the product of their slopes is \(-1\):
     \[
     \left( -\frac{b}{100} \right) \left( \frac{b}{200} \right) = -1
     \]
   - Simplify the equation:
     \[
     -\frac{b^2}{20000} = -1
     \]
     \[
     \frac{b^2}{20000} = 1
     \]
     \[
     b^2 = 20000
     \]
     \[
     b = \sqrt{20000} = 100\sqrt{2}
     \]

4. **Find the greatest integer less than \( AD \):**
   - Since \( AD = b \), we have:
     \[
     AD = 100\sqrt{2}
     \]
   - Approximate \( \sqrt{2} \approx 1.414 \):
     \[
     100\sqrt{2} \approx 100 \times 1.414 = 141.4
     \]
   - The greatest integer less than \( 141.4 \) is:
     \[
     \lfloor 141.4 \rfloor = 141
     \]

The final answer is \(\boxed{141}\)