Re: [中學] 難題
※ 引述《hau (小豪)》之銘言：
: https://imgur.com/a/dvJDA
: 如上圖
: 應有巧妙的解法......
A X B 設 XX'//AD, YY'//AB
AXD = a, BXY = b, CYD = c, DY'ZX' = t
Y' Z Y 因 (AXZY')(ZYCX') = (XBYZ)(Y'ZX'D)
D X' C 故 (2a-t)(2c-t) = 2b*t => (t-a-b-c)^2 = (a+b+c)^2-4ac
DXY = ABCD-a-b-c = (2a+2b+2c-t)-a-b-c = a+b+c-t
= √((a+b+c)^2-4ac)
原題所求 = √(12^2-60)=2√21
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