In a convex quadrilateral $ABCD$ it is given that $\angle{CAB} = 40^{\circ}, \angle{CAD} = 30^{\circ}, \angle{DBA} = 75^{\circ}$, and $\angle{DBC}=25^{\circ}$. Find $\angle{BDC}$.
1. Given the angles in the convex quadrilateral \(ABCD\):
   \[
   \angle CAB = 40^\circ, \quad \angle CAD = 30^\circ, \quad \angle DBA = 75^\circ, \quad \angle DBC = 25^\circ
   \]

2. We need to find \(\angle BDC\).

3. First, consider \(\triangle ABC\). Since \(\angle CAB = 40^\circ\) and \(\angle DBC = 25^\circ\), we can find \(\angle BAC\) and \(\angle BCA\):
   \[
   \angle BAC = \angle BCA = 40^\circ
   \]
   This makes \(\triangle ABC\) isosceles with \(AB = BC\).

4. Let the perpendicular from \(B\) to \(AC\) cut \(AD\) at \(E\). This means \(E\) is the foot of the perpendicular from \(B\) to \(AC\).

5. In \(\triangle AEC\), since \(\angle CAB = 40^\circ\) and \(\angle CAD = 30^\circ\), we have:
   \[
   \angle EAC = \angle CAB - \angle CAD = 40^\circ - 30^\circ = 10^\circ
   \]

6. Since \(\angle AEC\) is the external angle to \(\triangle AEC\), we have:
   \[
   \angle AEC = 180^\circ - \angle EAC = 180^\circ - 10^\circ = 170^\circ
   \]

7. However, this contradicts the earlier statement that \(\angle AEC = 120^\circ\). Therefore, we need to re-evaluate the configuration.

8. Given \(\angle DBA = 75^\circ\) and \(\angle DBC = 25^\circ\), we can find \(\angle ABC\):
   \[
   \angle ABC = \angle DBA - \angle DBC = 75^\circ - 25^\circ = 50^\circ
   \]

9. Since \(\angle BAC = 40^\circ\), we can find \(\angle BCA\):
   \[
   \angle BCA = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 40^\circ - 50^\circ = 90^\circ
   \]

10. Now, consider \(\triangle BEC\). Since \(\angle BCA = 90^\circ\), \(\angle BEC\) is:
    \[
    \angle BEC = 180^\circ - \angle BCA = 180^\circ - 90^\circ = 90^\circ
    \]

11. Since \(D\) is on the external bisector of \(\angle BEC\), \(\angle BDC\) is half of \(\angle BEC\):
    \[
    \angle BDC = \frac{1}{2} \times 90^\circ = 45^\circ
    \]

The final answer is \(\boxed{45^\circ}\).