如果学了向量,可以这样证.
以下形如AB的记号都表示向量," · "都表示点乘.
∵I是△AFG的垂心,有IG⊥AB,IF⊥AC,
∴(AG-AI)·AB = 0,(AF-AI)·AC = 0,
∴AI·BC = AI·(AC-AB) = AF·AC-AG·AB ①.
而∵H是△ADE的垂心,有EH⊥AB,DH⊥AC,
∴(AE-AH)·AB = 0,(AD-AH)·AC = 0,
∴AH·BC = AH·(AC-AB) = AD·AC-AE·AB ②.
又∵AF = BD,AG = CE,
∴AE = AC+CE = AC+AG,AD = AB+BD = AB+AF.
代入②式得AH·BC = AF·AC-AG·AB.
与①式相减得IH·BC = (AH-AI)·BC = 0,即IH⊥BC.
如果没学向量,但学了余弦定理,可以有本质上一样的证明,只是写起来麻烦一点.
先证一个引理:对于平面上两条线段AB与CD,成立AC²-BC² = AD²-BD²的充要条件是AB⊥CD.
充分性证明不难,把直线AB与CD的交点作出来,用勾股定理就行了.
必要性可以用同一法,过C,D分别作AB的垂线,由等式和勾股定理证明垂足重合.
∵I是△AFG的垂心,有IG⊥AF,IF⊥AG,即IG⊥AB,IF⊥AC,
∴AI²-BI² = AG²-BG²,AI²-CI² = AF²-CF²,
∴BI²-CI² = AF²-CF²-AG²+BG² = (BG²-AG²)-(CF²-AF²) ①.
由余弦定理,BG² = AG²+AB²-2AG·AB·cos∠BAG = AG²+AB²+2AG·AB·cos∠BAC,
CF² = AF²+AC²-2AF·AC·cos∠FAC = AF²+AC²+2AF·AC·cos∠BAC.
代入①式得BI²-CI² = AB²-AC²+2(AG·AB-AF·AC)cos∠BAC ②.
∵H是△ADE的垂心,有EH⊥AD,DH⊥AE,也即EH⊥AB,DH⊥AC.
∴AH²-BH² = AE²-BE²,AH²-CH² = AD²-CD²,
∴BH²-CH² = AD²-CD²-AE²+BE² = (BE²-AE²)-(CD²-AD²) ③.
由余弦定理,BE² = AE²+AB²-2AE·AB·cos∠BAC,CD² = AD²+AC²-2AD·AC·cos∠BAC.
代入③式得BH²-CH² = AB²-AC²-2(AE·AB-AD·AC)·cos∠BAC ④.
又∵AF = BD,AG = CE,
∴AE = AC-CE = AC-AG,AD = AB-BD = AB-AF,
∴AE·AB-AD·AC = -AG·AB+AF·AC.
代入④式并与②式比较得BI²-CI² = BH²-CH²,于是IH⊥BC.
