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HMMT 二月 2012 · TEAM1 赛 · 第 1 题

HMMT February 2012 — TEAM1 Round — Problem 1

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 20 ] Let ABC be a triangle. Let the angle bisector of ∠ A and the perpendicular bisector of BCintersect at D . Then let E and F be points on AB and AC such that DE and DF are perpendicularto AB and AC , respectively. Prove that BE = CF .
解析

英文解析

  1. [ 20 ] Let ABC be a triangle with AB < AC . Let the angle bisector of ∠ A and the perpendicularbisector of BC intersect at D . Then let E and F be points on AB and AC such that DE and DF areperpendicular to AB and AC , respectively. Prove that BE = CF .
    Answer: see below Note that DE, DF are the distances from D to AB, AC , respectively, andbecause AD is the angle bisector of ∠ BAC , we have DE = DF . Also, DB = DC because D is on the
    °
    ∼
    perpendicular bisector of BC . Finally, ∠ DEB = ∠ DF C = 90 , so it follows that DEB DF C , and
    BE = CF .=