Let $a,$ $b,$ $c,$ and $d$ be real numbers such that $a^2 + b^2 = 8$ and $c^2 + d^2 = 13.$  Find
\[(ad - bc)^2 + (ac + bd)^2.\]
To solve this problem, I need to find a way to relate the expressions $(ad - bc)^2 + (ac + bd)^2$ and $a^2 + b^2 = 8$ and $c^2 + d^2 = 13.$ I notice that the expression $(ad - bc)^2 + (ac + bd)^2$ can be rewritten as $(a^2 + b^2)(c^2 + d^2)$ by expanding and collecting like terms. This means that I can substitute $a^2 + b^2 = 8$ and $c^2 + d^2 = 13$ and get $(ad - bc)^2 + (ac + bd)^2 = (8)(13) = 104.$

