在△ABC中，如果cot(A/2), cot(B/2), cot(C/2) 成等差数列，则证明a, b, c成等差数列。

已知：在△ABC中，cot(A/2), cot(B/2), cot(C/2) 成等差数列。

求证：a, b, c成等差数列。

cot(A/2) = (S-a)/r

cot(B/2) = (S-b)/r

cot(C/2) = (S-c)/r

 

S = (a+b+c)/2

 

r = 半径

cot(A/2)/(S-a) = cot(B/2)/(S-b) = cot(C/2)/(S-c)

 

因为cot(A/2), cot(B/2), cot(C/2)成等差数列，

所以cot(B/2) = [cot(A/2) + cot(C/2)]/2

所以[cot(A/2) + cot(C/2)]/2 = {[(S-a)/(S-b)]cot(B/2) + [(S-c)/(S-b)]cot(B/2)}/2

 

= (1/2) * [(2S-a-c)/(S-b)] * cot(B/2)

= (1/2) * (a+2b+c)/(a+c) * cot(B/2)

= (1/2) * (2a+2c)/(a+c) * cot(B/2) [因为2b=a+c]

= cot(B/2)

所以a, b, c成等差数列。