Let $ABC$ be an acute angled triangle with $AB=15$ and $BC=8$. Let $D$ be a point on $AB$ such that $BD=BC$. Consider points $E$ on $AC$ such that $\angle DEB=\angle BEC$. If $\alpha$ denotes the product of all possible values of $AE$, find $\lfloor \alpha \rfloor$ the integer part of $\alpha$.
1. **Identify the given information and set up the problem:**
   - Triangle \(ABC\) is acute-angled.
   - \(AB = 15\) and \(BC = 8\).
   - Point \(D\) is on \(AB\) such that \(BD = BC = 8\).
   - Points \(E\) on \(AC\) such that \(\angle DEB = \angle BEC\).

2. **Determine the position of point \(D\):**
   - Since \(BD = BC = 8\), point \(D\) lies on \(AB\) such that \(BD = 8\).
   - Let \(AD = x\). Then \(AB = AD + DB = x + 8 = 15\).
   - Solving for \(x\), we get \(x = 7\).
   - Therefore, \(D\) divides \(AB\) into segments \(AD = 7\) and \(DB = 8\).

3. **Analyze the condition \(\angle DEB = \angle BEC\):**
   - This implies that \(E\) lies on the circumcircle of \(\triangle BDC\) because \(\angle DEB\) and \(\angle BEC\) are subtended by the same arc \(BC\).

4. **Use the properties of the circumcircle:**
   - Since \(D\) and \(C\) are fixed points, the circumcircle of \(\triangle BDC\) is fixed.
   - The point \(E\) must lie on the arc \(AC\) of this circumcircle.

5. **Calculate the possible values of \(AE\):**
   - Since \(E\) lies on the circumcircle, there are two possible positions for \(E\) on \(AC\) (one on each side of the line \(BC\)).
   - Let \(E_1\) and \(E_2\) be the two possible positions of \(E\).

6. **Determine the lengths \(AE_1\) and \(AE_2\):**
   - By symmetry and the properties of the circumcircle, the lengths \(AE_1\) and \(AE_2\) are equal.
   - Let \(AE_1 = AE_2 = AE\).

7. **Calculate the product of all possible values of \(AE\):**
   - Since both possible values of \(AE\) are equal, the product of all possible values of \(AE\) is \(AE \times AE = AE^2\).

8. **Find the integer part of \(\alpha\):**
   - Since \(AE\) is a fixed length, \(\alpha = AE^2\).
   - The problem asks for \(\lfloor \alpha \rfloor\), the integer part of \(\alpha\).

9. **Conclusion:**
   - The problem statement claims that there are infinitely many solutions, but this is incorrect. There are exactly two possible positions for \(E\), and both yield the same length \(AE\).
   - Therefore, the product of all possible values of \(AE\) is \(AE^2\), and we need to find \(\lfloor AE^2 \rfloor\).

The final answer is \( \boxed{ \lfloor AE^2 \rfloor } \).