So: a=kb, c=kd

1, difference: a+d-(b+c) = kb+d-b-KD = (k-1) b-(k-1) d = (k-1) (b-d) >.

So: a+d > b+c

2. Quotient: A AB BC DD C/A BB AC CD D.

=k^ab^(a+b)k^dd^(c+d)/k^bb^(a+b)k^cd^(c+d)

=k^(a+d)/k^(b+c)

= k (a+d-b-c), and the power of positive numbers greater than 1 is greater than 1.

So: a ab BC DD c/a bb AC CD d > 1

That is to say: from 2000 BC to 2000 BC, a bb AC CD D.