Is it possible that $\frac{|A \cap B|}{|A \cup B|} > \frac 12$ and $\frac{|A \cap C|}{|A \cup C|} > \frac 12$ given that $|B \cap C| = 0$? Given three sets, $A$, $B$, $C$, I have that  $B$ and $C$ are disjoint, and $|\cdot|$ represents the number of elements in the set:
$$|B \cap C| = 0$$ 
Is it possible that both:
$$\frac{|A \cap B|}{|A \cup B|} > \frac 12 $$
and
$$\frac{|A \cap C|}{|A \cup C|} > \frac 12 $$
are satisfied at the same time?

$\frac {|A \cap B|}{|A|} \ge  \frac {|A \cap B|}{|A \cup B|} > 1/2$ so more than half of the elements of A are in B.  Likewise more than half the elements of A are in C.  B and C can't both have more than 1/2 the elements and be distinct.  That'd means there'd be more then 100% of the elements accounted for.
