※ 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
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