HMMT 二月 2005 · 冲刺赛 · 第 26 题

HMMT February 2005 — Guts Round — Problem 26

[9] In triangle ABC , AC = 3 AB . Let AD bisect angle A with D lying on BC , and let 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

In triangle ABC , AC = 3 AB . Let AD bisect angle A with D lying on BC , and let E be 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 . But then 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. 9 C E D A B G F