Let $ABCD$ be a parallelogram that is not rhombus. We draw the symmetrical half-line of $(DC$ with respect to line $BD$. Similarly we draw the symmetrical half- line of $(AB$ with respect to $AC$. These half- lines intersect each other in $P$. If $\frac{AP}{DP}= q$ find the value of $\frac{AC}{BD}$ in function of $q$.
1. **Identify the given elements and their properties:**
   - \(ABCD\) is a parallelogram, which implies \(AB \parallel CD\) and \(AD \parallel BC\).
   - The symmetrical half-line of \(DC\) with respect to line \(BD\) is drawn.
   - The symmetrical half-line of \(AB\) with respect to line \(AC\) is drawn.
   - These half-lines intersect at point \(P\).
   - We need to find \(\frac{AC}{BD}\) in terms of \(q\), where \(q = \frac{AP}{DP}\).

2. **Use symmetry and properties of parallelograms:**
   - Since \(ABCD\) is a parallelogram, \(AC\) and \(BD\) are its diagonals.
   - The diagonals of a parallelogram bisect each other.

3. **Analyze the symmetrical half-lines:**
   - The symmetrical half-line of \(DC\) with respect to \(BD\) implies that the angle between \(DC\) and the symmetrical half-line is equal to the angle between \(BD\) and the symmetrical half-line.
   - Similarly, the symmetrical half-line of \(AB\) with respect to \(AC\) implies that the angle between \(AB\) and the symmetrical half-line is equal to the angle between \(AC\) and the symmetrical half-line.

4. **Use angle properties and bisectors:**
   - Let \(\alpha = m(\angle BAC)\) and \(\beta = m(\angle BDC)\).
   - In quadrilateral \(POCD\), we have:
     \[
     \alpha + \phi + \alpha + 2\beta + \phi = 360^\circ
     \]
     Simplifying, we get:
     \[
     2\alpha + 2\phi + 2\beta = 360^\circ \implies \alpha + \phi + \beta = 180^\circ
     \]

5. **Use similarity of triangles:**
   - Consider triangles \(\Delta AOP\) and \(\Delta ODP\).
   - Since \(\angle AOP = \beta\) and \(\angle POD = \alpha\), and \(\angle APO = \angle DPO\), the triangles \(\Delta AOP\) and \(\Delta ODP\) are similar by AA similarity criterion.
   - Therefore, we have:
     \[
     \frac{AP}{DP} = \frac{AO}{DO}
     \]
     Given \(\frac{AP}{DP} = q\), we get:
     \[
     q = \frac{AO}{DO}
     \]

6. **Relate \(AO\) and \(DO\) to \(AC\) and \(BD\):**
   - Since \(AC\) and \(BD\) are diagonals of the parallelogram and they bisect each other, we have:
     \[
     AO = \frac{AC}{2} \quad \text{and} \quad DO = \frac{BD}{2}
     \]
   - Therefore:
     \[
     q = \frac{AO}{DO} = \frac{\frac{AC}{2}}{\frac{BD}{2}} = \frac{AC}{BD}
     \]

7. **Solve for \(\frac{AC}{BD}\):**
   - From the above relation, we get:
     \[
     \frac{AC}{BD} = \sqrt{q}
     \]

The final answer is \(\boxed{\sqrt{q}}\).