证明: AC=b,AB=c 由余弦定理, cosC = [(p+q)2+b2-c2] / [2b(p+q)] = (b2+q2-AD2) / (2bq) 即 [(p+q)2+b2-c2] / (p+q) = (b2+q2-AD2) / q cosB = [(p+q)2+c2-b2] / [2c(p+q)] = (c2+p2-AD2) / (2cp) 即[(p+q)2+c2-b2] / (p+q) = (c2+p2-AD2) / p 两式相加得 p+q = (b2+q2-AD2) / q + (c2+p2-AD2) / p 解得 AD2 = (b2p+c2q) / (p+q)