The condition is a 2-b 2 = BC, where COSA = (B2+C2-A2)/2bc = (C2-BC)/2bc = (c-b)/2b,
cosb=(a^2+c^2-b^2)/2ac=(c^2+bc)/2ac=(c+b)/2a,
Then cos2b = 2 (cosb) 2-1= 2 * (c+b) 2/4a2-1= (c+b) 2/2a2-1,a 2 = b (b+c).
Cos2b = (c+b)/2b-1= (c-b)/2b = COSA, A and B are both interior angles of a triangle, and A=2B.
2. Conversely, it is easy to prove. If A=2B, according to sine theorem:
SinA=asinB/b=sin2B=2sinBcosB, and cosB=a/2b combined with cosine theorem, then
Cosb = (a 2+c 2-b 2)/2ac = a/2b, which is simplified as a 2 = b (b+c).
So the relationship between the two is "sufficient and necessary", and the answer is A.