(1)、【探索发现】如图1,△ABC中,点D,E,F分别在边BC,AC,AB上,且AD,BE,CF相交于同一点O.用”S”表示三角形的面积,有S
△ABD:S
△ACD=BD:CD,这一结论可通过以下推理得到:过点B作BM⊥AD,交AD延长线于点M,过点C作CN⊥AD于点N,可得S
△ABD:S
△ACD=
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmo+stretchy%3D%22false%22%3E%28%3C%2Fmo%3E%3Cmfrac%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmfrac%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EO%3C%2Fmi%3E%3Cmo%3E%E2%8B%85%3C%2Fmo%3E%3Cmi%3EB%3C%2Fmi%3E%3Cmi%3EM%3C%2Fmi%3E%3Cmo+stretchy%3D%22false%22%3E%29%3C%2Fmo%3E%3Cmo%3E%EF%BC%9A%3C%2Fmo%3E%3Cmo+stretchy%3D%22false%22%3E%28%3C%2Fmo%3E%3Cmfrac%3E%3Cmn%3E1%3C%2Fmn%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmfrac%3E%3Cmi%3EA%3C%2Fmi%3E%3Cmi%3EO%3C%2Fmi%3E%3Cmo%3E%C2%B7%3C%2Fmo%3E%3Cmi%3EC%3C%2Fmi%3E%3Cmi%3EN%3C%2Fmi%3E%3Cmo+stretchy%3D%22false%22%3E%29%3C%2Fmo%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
,又可证△BDM~△CDN,∴BM:CN=BD:CD,∴S
△ABD:S
△ACD=BD:CD.由此可得S
△BAO:S
△BCO=
;S
△CAO:S
△CBO=
;若D,E,F分别是BC,AC,AB的中点,则S
△BFO:S
△ABC=
.