已知△ABC中,AD是角BAC的角平分线,求证AB/AC=BC/DC.(探索多种证明方法)起码要5种哦,明天早上7点钟以前给我答案,

Question

已知△ABC中,AD是角BAC的角平分线,求证AB/AC=BC/DC.(探索多种证明方法)
起码要5种哦,明天早上7点钟以前给我答案,

Answer 1

结论有误,应该是:AB/AC=BD/DC.
证法1:作DE平行CA,交AB于E,则⊿BED∽⊿BAC;且∠EDA=∠DAC;
又∠EAD=∠DAC,故∠EAD=∠EDA,AE=DE.
故:AB/AC=BE/DE=BE/AE=BD/DC.
证法2:作DE平行AB,交AC于E.
同理可证:DE=AE;⊿ABC∽⊿EDC.
则AB/AC=ED/EC=AE/EC=BD/DC.
证法3:作CE平行AD,交BA的延长线于E.
则⊿BAD∽⊿BEC;且∠E=∠BAD;∠ACE=∠CAD.
又∠BAD=∠CAD.故∠E=∠ACE,AC=AE.
所以,AB/AC=AB/AE=BD/DC.
证法4:过点C作AB的平行线,交AD的延长线于E.
则⊿ABD∽⊿ECD;且∠E=∠BAD=∠CAD,CE=AC.
故:AB/AC=AB/CE=BD/DC.
证法5:作DE垂直AB于E,DF垂直AC于F.AD平分∠BAC,则DE=DF.
故S⊿ABD/S⊿ACD=(AB*DE/2)/(AC*DF/2)=AB/AC;
又S⊿ABD/S⊿ACD=BD/DC.(同高的三角形面积比等于底之比)
所以,AB/AC=BD/DC.