If integers $a$, $b$, $c$, and $d$ satisfy $ bc + ad = ac + 2bd = 1 $, find all possible values of $ \frac {a^2 + c^2}{b^2 + d^2} $.
1. We start with the given equations:
   \[
   bc + ad = 1
   \]
   \[
   ac + 2bd = 1
   \]

2. Subtract the first equation from the second:
   \[
   ac + 2bd - (bc + ad) = 1 - 1
   \]
   Simplifying, we get:
   \[
   ac + 2bd - bc - ad = 0
   \]
   \[
   ac - bc + 2bd - ad = 0
   \]
   Factor out common terms:
   \[
   c(a - b) + d(2b - a) = 0
   \]

3. Rearrange the equation:
   \[
   c(a - b) = d(a - 2b)
   \]
   If \(a \neq b\), we can divide both sides by \(a - b\):
   \[
   c = \frac{d(a - 2b)}{a - b}
   \]

4. Now, consider the greatest common divisor (gcd) conditions:
   \[
   \gcd(a, b) \mid bc \quad \text{and} \quad \gcd(a, b) \mid ad
   \]
   Since \(\gcd(a, b) \mid bc + ad = 1\), it follows that \(\gcd(a, b) = 1\).

5. Similarly, \(\gcd(c, d) = 1\). Now, consider the gcd of \(b - a\) and \(b\):
   \[
   \gcd(b - a, b) = \gcd(b - a, a) = 1
   \]
   Since \((b - a) \mid bd\), we must have \((b - a) \mid d\).

6. Similarly, \(\gcd(c - d, d) = 1\) and \((c - d) \mid b\).

7. Given \((b - a) \mid d\) and \((c - d) \mid b\), and knowing \((b - a)(c - d) = bd\), we have two cases:
   - \(b - a = d\) and \(c - d = b\)
   - \(b - a = -d\) and \(c - d = -b\)

8. For the first case:
   \[
   a = b - d \quad \text{and} \quad c = b + d
   \]
   For the second case:
   \[
   a = b + d \quad \text{and} \quad c = d - b
   \]

9. In both cases, we calculate \(a^2 + c^2\):
   - First case:
     \[
     a^2 + c^2 = (b - d)^2 + (b + d)^2 = b^2 - 2bd + d^2 + b^2 + 2bd + d^2 = 2b^2 + 2d^2
     \]
   - Second case:
     \[
     a^2 + c^2 = (b + d)^2 + (d - b)^2 = b^2 + 2bd + d^2 + d^2 - 2bd + b^2 = 2b^2 + 2d^2
     \]

10. Therefore, in both cases:
    \[
    \frac{a^2 + c^2}{b^2 + d^2} = \frac{2b^2 + 2d^2}{b^2 + d^2} = 2
    \]

Conclusion:
\[
\boxed{2}
\]