A quadrilateral $ABCD$ has a right angle at $\angle ABC$ and satisfies $AB = 12$, $BC = 9$, $CD = 20$, and $DA = 25$. Determine $BD^2$.
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1. Given the quadrilateral \(ABCD\) with a right angle at \(\angle ABC\), we know:
   \[
   AB = 12, \quad BC = 9, \quad CD = 20, \quad DA = 25
   \]
   We need to determine \(BD^2\).

2. Let \(\angle BCA = \theta\). Since \(\triangle ABC\) is a right triangle at \(B\), we can use trigonometric identities to find \(\sin \theta\) and \(\cos \theta\).

3. In \(\triangle ABC\):
   \[
   \sin \theta = \frac{AB}{AC} = \frac{12}{15} = \frac{4}{5}
   \]
   \[
   \cos \theta = \frac{BC}{AC} = \frac{9}{15} = \frac{3}{5}
   \]
   where \(AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\).

4. By the Law of Cosines on \(\triangle BCD\):
   \[
   BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos(\angle BCD)
   \]
   Since \(\angle BCD = 90^\circ + \theta\), we use the identity \(\cos(90^\circ + \theta) = -\sin \theta\):
   \[
   \cos(90^\circ + \theta) = -\sin \theta = -\frac{4}{5}
   \]

5. Substituting the values into the Law of Cosines formula:
   \[
   BD^2 = 9^2 + 20^2 - 2 \cdot 9 \cdot 20 \cdot \left(-\frac{4}{5}\right)
   \]
   \[
   BD^2 = 81 + 400 + 2 \cdot 9 \cdot 20 \cdot \frac{4}{5}
   \]
   \[
   BD^2 = 481 + 2 \cdot 9 \cdot 4 \cdot 4
   \]
   \[
   BD^2 = 481 + 288
   \]
   \[
   BD^2 = 769
   \]

The final answer is \(\boxed{769}\).