题目

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

提问时间：2021-03-26

答案

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