MQ31 --- Triangle

圖片參考：http://imgcld.yimg.com/8/n/HA04628698/o/701207120054913873412470.jpg

△XBC / △XBA = CQ / QA
[△XBP (n+1)/n] / △XBA = n
△XBP / △XBA = n² / (n+1)Let △XBP = n² = △YCQ = △ZAR
Then △XBA = n+1 = △YCB = △ZAC ∴
∆ABC = (∆XYZ + △XBA + △YCB + △ZAC) = ∆XYZ + 3(n+1)
On the other hand ,
∆ABC = (△XBA + △XBP) (n+1)/n = (n² + n+1) (n+1)/n ∴
∆ABC : ∆XYZ
= (n²+n+1)(n+1)/n : [(n²+n+1)(n+1)/n - 3(n+1)]
= 1 : [1 - 3n/(n²+n+1)]
= 1 : [(n² -2n+1)/(n²+n+1)]
= (n² + n + 1) / (n - 1)²

2012-07-13 23:28:00 補充：
Similarly , △YCQ / △YCB = n² / (n+1) ,

therefore △YCQ / △BCQ = △XBP / △ABP ,

but △ABP = △BCQ = △ABC * n/(n+1) ,

thus △YCQ = △XBP.

Similarly , △XBP = △YCQ = △ZAR.