To the contrary, assume that $ac + bd$ is prime. Then $ac + bd \mid ab + cd \Rightarrow ac + bd \mid ac + bd + ab + cd = (a + d)(b + c)$, and so $ac + bd \mid a + d$ or $ac + bd \mid b + c$. But this is impossible because for distinct $a, b, c, d$ we have that $ac + bd > a + d$ and $ac + bd > b + c$. This contradiction completes the proof.