Let $ABCD$ be a convex quadrilateral with $AB=AD, m\angle A = 40^{\circ}, m\angle C = 130^{\circ},$ and $m\angle ADC - m\angle ABC = 20^{\circ}.$ Find the measure of the non-reflex angle $\angle CDB$ in degrees. 
1. Given that \(ABCD\) is a convex quadrilateral with \(AB = AD\), we know that \(\triangle ABD\) is isosceles with \(AB = AD\).
2. We are given \(m\angle A = 40^\circ\). Since \(\triangle ABD\) is isosceles, the base angles \(\angle ABD\) and \(\angle ADB\) are equal. Let these angles be denoted as \(\alpha\).
3. Using the fact that the sum of the angles in a triangle is \(180^\circ\), we have:
   \[
   \angle A + \angle ABD + \angle ADB = 180^\circ
   \]
   Substituting the known values:
   \[
   40^\circ + \alpha + \alpha = 180^\circ
   \]
   Simplifying, we get:
   \[
   40^\circ + 2\alpha = 180^\circ
   \]
   \[
   2\alpha = 140^\circ
   \]
   \[
   \alpha = 70^\circ
   \]
   Therefore, \(\angle ABD = \angle ADB = 70^\circ\).

4. We are also given \(m\angle C = 130^\circ\) and \(m\angle ADC - m\angle ABC = 20^\circ\). Let \(\angle ABC = \beta\) and \(\angle ADC = \gamma\).
5. From the given condition, we have:
   \[
   \gamma - \beta = 20^\circ
   \]

6. Since \(\angle ABD = 70^\circ\), we can denote \(\angle DBC = x\). Then, \(\angle BDC = 50^\circ - x\) because the sum of angles around point \(D\) must be \(180^\circ\) (since \(\angle ADB = 70^\circ\) and \(\angle ADC = \gamma\)).

7. Using the given condition \(\gamma - \beta = 20^\circ\), we substitute \(\gamma = 50^\circ - x\) and \(\beta = x\):
   \[
   (50^\circ - x) - x = 20^\circ
   \]
   Simplifying, we get:
   \[
   50^\circ - 2x = 20^\circ
   \]
   \[
   30^\circ = 2x
   \]
   \[
   x = 15^\circ
   \]

8. Therefore, \(\angle DBC = 15^\circ\) and \(\angle BDC = 50^\circ - 15^\circ = 35^\circ\).

The final answer is \(\boxed{35^\circ}\).