作者Honor1984 (奈何上天造化弄人？)
看板Math
标题Re: [几何] 请问一题国中几何
时间Tue May 17 17:20:36 2022
※ 引述《jenshi (小旭)》之铭言：
: https://imgur.com/a/6ocu1pq
: 是否有办法证明ADE是等腰三角形
令AD = 2x，，AB = 2y
C、B的垂足分别为C'、B'，BB'交CC'於H
DH = tx，HB = ty，B'D = x
=> C'H = sqrt(t^2 - 1)y = uy，u = qsrt(t^2 - 1)
t^2 - u^2 = 1
AC = ty/u
CC' = y/u
AB' = 2uy/t
CB' = ty/u - 2uy/t = [y/(tu)][t^2 - 2u^2] = [y/(tu)][1 - u^2]
CH = y/u - uy = (y/u)[1 - u^2]
=> CB' : CH = 1 : t = B'D : DH
=> CE为角ACC'的分角线
=> 角DEB = 80 = 角BDE
=> 三角形BDE为等腰三角形
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1F：→ Honor1984 : 接着也就可求出∠ADE=40 => △EAD为等腰三角形 05/18 03:17