Question

三角形内角平分线性质定理：三角形的内角平分线分对边所得的两条线段和这个角的两边对应成比例。 已知：如图，△ABC中，AD是角平分线． 求证：(1)BD/DC＝AB/AC (2)若AD是三角形ABC外角的平分线,交BC延长线于点D，是否还有以上结论？

Answer 1

(1) 过C作CE∥DA，交BA的延长线于E．∵CE∥DA∴∠1＝∠E，∠2＝∠3，∠1＝∠2∴∠E＝∠3∴AE＝AC∵CE∥DA∴BD/DC＝BA/AE又∵AE＝AC∴BD/DC＝AB/AC(2)在BA延长线上取点C'，使AC'=AC，过C'作C'D'//CD交DA延长线于点D',连接C'D.∵C'D'//CD，A是BA与DD'的交点∴△ABD∽△AC'D'∴BD/C'D'=AB/AC'∵C'D'//CD∴∠C'D'A=∠ADB∵AD是三角形ABC外角的平分线∴∠C'AD=∠CAD∵AC'=AC,AD是公共边∴△C'AD≌△CAD∴∠C'DA=∠ADB,C'D=CD∴∠C'DA=∠C'D'A∴C'D'=C'D=CD∴BD/DC=BD/C'D'=AB/AC'=AB/AC