Given 3 disjoint Turing-recognizable languages prove that one of them is decidable Let $A$, $B$, $C$ be Turing-recognizable languages over an alphabet $\Sigma$. Assume that

$A\cup B\cup C = \Sigma^*$, and
$A\cap B = A\cap C = B\cap C = \emptyset$.

Prove that $A$ is Turing-decidable.

So far I've tried the following but it really hasn't gotten me anywhere.
Suppose $A$ is recognizable but not decidable then, $\overline{A} = B \cup C$ so $B$ or $C$ is undecidable, since decidable languages are closed under union.
In the case that only one is decidable you can show at least that its in $A$ or $B$ (or $A$ or $C$ in the other case). Then, ... ?
In the case that both are recognizable, ... ?
Let's call MA, MB and MC the Turing machines that recognize A, B and C, respectively. The 3-TM T (a Turing machine with 3 tapes) that decides A can be described as follows:

T: input x. 


Copy x on 2nd and 3rd tapes.
T executes one step of MA on tape 1. If MA accepts x, T accepts x.
T executes one step of MB on tape 2. If MB accepts x, T rejects x.
T executes one step of MC on tape 3. If MC accepts x, T rejects x.


T halts for every string x in A U B U C, so T halts for every input in Σ*. Moreover, L(T) = A. So T is a TM that decides A, and A is Turing-decidable.