HMMT 二月 2005 · 冲刺赛 · 第 26 题
HMMT February 2005 — Guts Round — Problem 26
题目详情
英文原题
- [9] In triangle ABC , AC = 3 AB . Let AD bisect angle A with D lying on BC , andlet E be the foot of the perpendicular from C to AD . Find [ ABD ] / [ CDE ]. (Here,
[ XY Z ] denotes the area of triangle XY Z ).
解析
英文解析
- In triangle ABC , AC = 3 AB . Let AD bisect angle A with D lying on BC , and let Ebe the foot of the perpendicular from C to AD . Find [ ABD ] / [ CDE ]. (Here, [ XY Z ]
denotes the area of triangle XY Z ).
Solution: 1 / 3
By the Angle Bisector Theorem, DC/DB = AC/AB = 3. We will show that AD =
DE . Let CE intersect AB at F . Then since AE bisects angle A , AF = AC = 3 AB ,
and EF = EC . Let G be the midpoint of BF . Then BG = GF , so GE ‖ BC . Butthen since B is the midpoint of AG , D must be the midpoint of AE , as desired. Then
[ ABD ] / [ CDE ] = ( AD · BD ) / ( ED · CD ) = 1 / 3.
C9
A B G FDE