如图,D为三角形ABC的边BC上一点,DE平行AB,DF平行AC分别交AC,AB于E,F,设三角形CDE,

证明:DE∥AB,则:?CDE∽?ABC,S?CDE/S?ABC=(CD./BD)^2. 即:S1/(S1+S2+S3)=(CD/BC)^2,√[S1/(S1+S2+S3)]=CD/BC;---------------------(1) 同理可证:√[S2/(S1+S2+S3)]=BD/BC;------------------------------------------------(2) ∴√[S1/(S1+S2+S3)]