d + e < a + b and a + b < c + d implies by transitivity  of < that

d + e < c + d

Subtract d from both sides

e < c

which is choice A).

But we must rule out the fact that none of the other is ALWAYS
true.

We can rule out B), C) and D) with just one case

a=5, b=1, c=5, d=2, e=2

5 1 5 2 2   b) c) d)

A)less than c   B) equal to c   C) less than d   D) greater than d

 a + b  is less than c + d , and d + e is less than a + b 
 5 + 1  is less than 5 + 2 , and 2 + 2 is less than 5 + 1
   6    is less than   7   , and   4   is less than   6

yet B) is not true since e=2 is not equal to c=5

also C) is not true since e=2 is not less than d=2

also D) is not true since e=2 is not greater than d=2 

In fact, it can be shown that B) and D) are NEVER true, 

and that C) is only sometimes true.  

Edwin