若AD是三角形ABC外角的平分线,交BC延长线于点D,试根据相似三角形对应边成比例说明BD/DC=AB/AC

Question

如题所述

Answer 1

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

Answer 2

俊狼猎英团队为您解答

过D作DE∥AC交BA的延长线于E，
则ΔBAC∽ΔBED，∠ADE=∠CAD
∴BE/BA=BD/BC……① AB/AC=BE/DE，……③
∴BE/BA-1=BD/BC-1
(BE-BA)/BA=(BD-BC)/BC
即AE/AB=CD/BC……②
①÷②得：BE/AE=BD/CD……④
∵AD是∠DAE的平分线，∴∠CAD=∠EAD，
∴∠EAD=∠ADE，∴AE=DE，代入③
得AB/AC=BE/AE，与④结合得：
AB/AC=BD/CD本回答被网友采纳