In his triangle $ABC$ Serge made some measurements and informed Ilias about the lengths of median $AD$ and side $AC$. Based on these data Ilias proved the assertion: angle $CAB$ is obtuse, while angle $DAB$ is acute. Determine a ratio $AD/AC$ and prove Ilias' assertion (for any triangle with such a ratio).
1. **Understanding the Problem:**
   We are given a triangle \(ABC\) with median \(AD\) and side \(AC\). We need to determine the ratio \( \frac{AD}{AC} \) and prove that \(\angle CAB\) is obtuse while \(\angle DAB\) is acute.

2. **Using the Median Property:**
   Recall that the median of a triangle divides it into two smaller triangles of equal area. Also, the median \(AD\) from vertex \(A\) to the midpoint \(D\) of side \(BC\) can be expressed using Apollonius's theorem:
   \[
   AB^2 + AC^2 = 2AD^2 + 2BD^2
   \]
   Since \(D\) is the midpoint of \(BC\), \(BD = DC = \frac{BC}{2}\).

3. **Condition for Obtuse Angle:**
   \(\angle CAB\) is obtuse if and only if \(A\) lies outside the circle with diameter \(BC\). This implies that \(AD > DC\).

4. **Inequality Analysis:**
   If \(AC < 2AD\), then there exists a triangle such that \(AD = DC\), making \(\angle CAB\) a right angle. Therefore, for \(\angle CAB\) to be obtuse, we must have:
   \[
   AC \geq 2AD
   \]

5. **Midpoint Consideration:**
   Let \(E\) be the midpoint of \(AC\). Then, \(\angle BAD = \angle ADE\). If \(AC > 2AD\), then \(AE > AD\), and there exists a triangle such that \(\angle ADE = 90^\circ\), making \(\angle BAD\) not acute. Therefore, for \(\angle DAB\) to be acute, we must have:
   \[
   AC \leq 2AD
   \]

6. **Combining Inequalities:**
   Combining the inequalities \(AC \geq 2AD\) and \(AC \leq 2AD\), we get:
   \[
   AC = 2AD
   \]

7. **Verification:**
   To verify, consider any triangle with \( \frac{AD}{AC} = \frac{1}{2} \). This implies \(AC = 2AD\). Using the properties of medians and the conditions derived, it can be shown that \(\angle CAB\) is obtuse and \(\angle DAB\) is acute.

\(\blacksquare\)

The final answer is \( \boxed{ \frac{AD}{AC} = \frac{1}{2} } \)