If $A+B>C+D$ and $B+C>A+D$, then it is definite that

We are given two inequalities:
1. $A+B>C+D$
2. $B+C>A+D$

Let us add inequality (1) and inequality (2) together:

$$(A+B)+(B+C)>(C+D)+(A+D)$$

$$A+2B+C>A+C+2D$$

Subtracting $A$ and $C$ from both sides:

$$2B>2D$$

$$B>D$$

Hence, it is definitive that $B>D$.