In trapezoid $ABCD$, $AD \parallel BC$ and $\angle ABC + \angle CDA = 270^{\circ}$.  Compute $AB^2$ given that $AB \cdot \tan(\angle BCD) = 20$ and $CD = 13$.

[i]Proposed by Lewis Chen[/i]
1. Given that in trapezoid \(ABCD\), \(AD \parallel BC\) and \(\angle ABC + \angle CDA = 270^\circ\). We need to compute \(AB^2\) given that \(AB \cdot \tan(\angle BCD) = 20\) and \(CD = 13\).

2. Since \(\angle ABC + \angle CDA = 270^\circ\), and knowing that \(\angle ABC > 90^\circ\) because \(\angle CDA < 180^\circ\), we can infer that \(\angle ABC\) is an obtuse angle.

3. Drop the altitude from \(D\) to \(BC\) and call the point \(G\), letting \(DG = a\) and \(CG = b\). Also, drop the altitude \(AH\) from \(A\) to \(BC\).

4. Since \(\angle BAH = \angle BCD\), triangles \(\triangle BAH\) and \(\triangle BCD\) are similar by AA similarity criterion.

5. From the similarity of \(\triangle BAH\) and \(\triangle BCD\), we have the ratio of corresponding sides:
   \[
   \frac{AB}{CD} = \frac{AH}{DG}
   \]
   Given \(CD = 13\), we can write:
   \[
   AB = \frac{13 \cdot AH}{DG}
   \]

6. Since \(\tan(\angle BCD) = \frac{DG}{CG} = \frac{a}{b}\), we can substitute \(DG = a\) and \(CG = b\) into the equation:
   \[
   AB \cdot \tan(\angle BCD) = 20
   \]
   Substituting \(\tan(\angle BCD) = \frac{a}{b}\), we get:
   \[
   AB \cdot \frac{a}{b} = 20
   \]

7. From the similarity, we also have:
   \[
   AB = \frac{13a}{b}
   \]

8. Substituting \(AB = \frac{13a}{b}\) into the equation \(AB \cdot \frac{a}{b} = 20\), we get:
   \[
   \frac{13a}{b} \cdot \frac{a}{b} = 20
   \]
   Simplifying, we get:
   \[
   \frac{13a^2}{b^2} = 20
   \]

9. Multiplying both sides by \(b^2\), we get:
   \[
   13a^2 = 20b^2
   \]

10. Solving for \(a^2\), we get:
    \[
    a^2 = \frac{20b^2}{13}
    \]

11. Substituting \(a^2\) back into the expression for \(AB^2\), we get:
    \[
    AB^2 = \left(\frac{13a}{b}\right)^2 = \frac{169a^2}{b^2}
    \]

12. Substituting \(a^2 = \frac{20b^2}{13}\) into the equation, we get:
    \[
    AB^2 = \frac{169 \cdot \frac{20b^2}{13}}{b^2} = \frac{169 \cdot 20}{13} = 260
    \]

The final answer is \(\boxed{260}\).