在△ABC中,三边的长分别为a,b,c,角A,B,C上的平分线分别记为na,nb,nc,应用余弦定理求na,nb,nc,

设角A的平分线交BC边于D点,由角平分线定理AB/AC = DB/DC知：
DB = a * (c/(b+c)) = ac/(b+c),DC = ab/(b+c)
在三角形ADB与三角形ADC中对∠ADB与∠ADC用余弦定理,注意到
∠ADB + ∠ADC = 180°
所以
0 = cos∠ADB + cos∠ADC = [na^2 + [ac/(b+c)]^2 - c^2] / (2 * na * ac/(b+c)) + [na^2 + [ab/(b+c)]^2 - b^2]/(2 * na * ab/(b+c))
这里把分母的na约去,是关于na^2的一次方程,通分（但不要全部展开）得：
(b+c)na^2 + b[(ac/(b+c))^2 - c^2] + c[(bc/(b+c))^2 - b^2]
= (b+c)na^2 + bc^2[(a/(b+c)]^2 - 1] + cb^2[(a/(b+c))^2 - 1]
= (b+c)na^2 + bc(b+c)[(a/(b+c)]^2 - 1]
= 0
所以na^2 = bc[1-(a/(b+c))^2] = bc/(b+c)^2(b+c-a)(b+c+a)
如果记半周长p = (a+b+c)/2
那么有na = 2/(b+c)√[bcp(p-a)]
同理有nb = 2/(c+a)√[cap(p-b)]
nc = 2/(a+b)√[abp(p-c)]