Show that $a \mid (b \times c) \not\!\!\!\implies a | b$ or $a \mid c$.
We can show this with a counter-example.
We take $a=10, b=6, c=15$, which gives us $10 \mid 90$ which is true, but both $10 \mid 6$ and $10 \mid 15$ are false.
If $b$ and $c$ are co-prime, that is, they share no common factors, then only one of $a \mid b$ and $a \mid c$ is true, as long as $a > 1$.