Let $ABCD$ be a cyclic quadrilateral such that the lines $AB$ and $CD$ intersects in $K$, let $M$ and $N$ be the midpoints of $AC$ and $CK$ respectively. Find the possible value(s) of $\angle ADC$ if the quadrilateral $MBND$ is cyclic.
1. Let \(ABCD\) be a cyclic quadrilateral with \(AB\) and \(CD\) intersecting at \(K\). Let \(M\) and \(N\) be the midpoints of \(AC\) and \(CK\) respectively. We need to find the possible value(s) of \(\angle ADC\) if the quadrilateral \(MBND\) is cyclic.

2. We will use complex numbers to represent the points on the unit circle. Let \(a, b, c, d\) be the complex numbers corresponding to points \(A, B, C, D\) respectively, with \(c = 1\) since \(C\) is on the unit circle.

3. Since \(K, C, D\) are collinear, we have:
   \[
   \frac{k-1}{\overline{k}-1} = \frac{d-1}{\overline{d}-1} = -d
   \]
   Solving for \(\overline{k}\), we get:
   \[
   \overline{k} = \frac{d+1-k}{d} \quad \text{(Equation I)}
   \]

4. Since \(K, B, A\) are collinear, we have:
   \[
   \frac{k-b}{\overline{k}-\overline{b}} = \frac{a-b}{\overline{a}-\overline{b}} = -ab
   \]
   Solving for \(\overline{k}\), we get:
   \[
   \overline{k} = \frac{a+b-k}{ab} \quad \text{(Equation II)}
   \]

5. Equating the two expressions for \(\overline{k}\) from Equations I and II, we get:
   \[
   \frac{d+1-k}{d} = \frac{a+b-k}{ab}
   \]
   Solving for \(k\), we get:
   \[
   k = \frac{abd + ab - ad - bd}{ab - d} \quad \text{(Equation III)}
   \]

6. The midpoint \(M\) of \(AC\) is:
   \[
   m = \frac{a + 1}{2}
   \]

7. The midpoint \(N\) of \(CK\) is:
   \[
   n = \frac{k + 1}{2}
   \]

8. For \(MBND\) to be cyclic, the following condition must hold:
   \[
   \frac{(m-n)(b-d)}{(b-n)(d-m)} \in \mathbb{R}
   \]

9. Substituting the values of \(m\) and \(n\), we get:
   \[
   \frac{\left(\frac{a+1}{2} - \frac{k+1}{2}\right)(b-d)}{\left(b - \frac{k+1}{2}\right)\left(d - \frac{a+1}{2}\right)} \in \mathbb{R}
   \]
   Simplifying, we get:
   \[
   \frac{(a-k)(b-d)}{(2b-k-1)(2d-a-1)} \in \mathbb{R}
   \]

10. Substituting the value of \(k\) from Equation III, we get:
    \[
    \frac{\left(a - \frac{abd + ab - ad - bd}{ab - d}\right)(b-d)}{(2b - \frac{abd + ab - ad - bd}{ab - d} - 1)(2d - a - 1)} \in \mathbb{R}
    \]

11. Simplifying the above expression, we get:
    \[
    \frac{b(a-1)(a-d)(b-d)}{(b-1)(2ab-ad-d)(2d-a-1)} \in \mathbb{R}
    \]

12. For the above expression to be real, the numerator and denominator must be conjugates. This implies:
    \[
    b(a-1)(a-d)(b-d) = (b-1)(2ab-ad-d)(2d-a-1)
    \]

13. Simplifying further, we get:
    \[
    b(a+1)(2ab+2d^2-bd-abd) = (a+1)(2d^2+2ab-ad-d)
    \]

14. If \(a \neq -1\), we get:
    \[
    b(2ab+2d^2-bd-abd) = 2d^2+2ab-ad-d
    \]
    Simplifying, we get:
    \[
    (b-1)(2ab+2d^2-bd-d-abd-ad) = 0
    \]
    \[
    2ab+2d^2-bd-d-abd-ad = 0
    \]
    \[
    a = \frac{bd+d-2d^2}{2b-bd-d} = \frac{(b+1-2d)d}{2b-bd-d}
    \]

15. For the above to hold, we must have:
    \[
    \frac{(b+1-2d)d}{2b-bd-d} = \frac{1}{\overline{\frac{(b+1-2d)d}{2b-bd-d}}}
    \]
    Simplifying, we get:
    \[
    d + bd - 2b = bd - 2b - d
    \]
    \[
    d = 0 \quad (\Rightarrow \Leftarrow)
    \]

16. Therefore, \(a = -1\), which implies \(AC\) is a diameter of the circle. Hence, \(\angle ADC = 90^\circ\).

The final answer is \(\boxed{90^\circ}\)