HMMT 二月 2012 · TEAM2 赛 · 第 6 题

HMMT February 2012 — TEAM2 Round — Problem 6

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

解析

[ 20 ] Let ABC be a triangle with AB < AC . Let the angle bisector of ∠ A and the perpendicular bisector of BC intersect at D . Then let E and F be points on AB and AC such that DE and DF are perpendicular 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, and because 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 .