Re: [幾何] 三角形內切圓證明

看板Math作者 (奈何上天造化弄人?)時間4年前 (2021/10/05 15:19), 編輯推噓0(000)
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※ 引述《goodwang (手牽手一起去奔跑)》之銘言: : https://i.imgur.com/DjSUNZJ.jpg
: 如圖片,要證明線段AR等於線段CQ : 感謝各位 B過O交AC於W,不難證明出 RW : WQ = (a + b + c) : (a - b + c) bc/(a + c) = AR + RW (b + c - a)/2 = WQ + bc/(a + c) => AR = bc/(a + c) - [(a + b + c)/(a - b + c)][(b + c - a)/2 - bc/(a + c)] = 2bc/(a - b + c) - (a + b + c)(b + c - a)/[2(a - b + c)] = [4bc - (b + c)^2 + a^2]/[2(a - b + c)] = (a + b - c)/2 = CQ -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 111.243.48.34 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1633418391.A.95E.html
文章代碼(AID): #1XM_oNbU (Math)
文章代碼(AID): #1XM_oNbU (Math)