![](https://i.imgur.com/nl750yL.jpg)
Re: [中學] 國中幾何證明題
![](https://i.imgur.com/nl750yL.jpg)
: 科學班甄選試題,求解
(對應的垂足有誤,以下過程的座標以題目為主)
FD BD
由四點共圓可知,∠DFB = ∠C,配上正弦定理 ------ = ------
sin∠B sin∠C
且 BD = AB*cos∠B = 2Rsin∠C*cos∠B
所以,FD = 2Rsin∠B*cos∠B = AC*cos∠B
(以下令 BC = a 、 CA = b 、AB = c )
代換原式 (DE+EF+FD)*R/2 = ( a*cos∠A + b*cos∠B + c*cos∠C )*R/2 --- (1)
注意到外接圓圓心O到三邊之距分別就是 Rcos∠A、Rcos∠B、Rcos∠C
所以(1)式就代表△OBC+△OCA+△OAB = △ ABC 證畢
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 124.11.128.7
※ 文章網址: http://www.ptt.cc/bbs/Math/M.1417187221.A.3CB.html
推
11/28 23:36, , 1F
11/28 23:36, 1F
→
11/28 23:36, , 2F
11/28 23:36, 2F
→
11/28 23:40, , 3F
11/28 23:40, 3F
推
11/28 23:41, , 4F
11/28 23:41, 4F
→
11/28 23:54, , 5F
11/28 23:54, 5F
OA = OB = OC = R
∠AOB = 2∠C (圓周角圓心角)
所以對半的垂線OM(假設交AB於M) ∠MOA = ∠MOB = ∠C
那OM自然就是RcosC
→
11/28 23:58, , 6F
11/28 23:58, 6F
→
11/29 00:15, , 7F
11/29 00:15, 7F
推
11/29 00:18, , 8F
11/29 00:18, 8F
ACDF四點共圓 ∠BFD = ∠C 不是FDB
※ 編輯: FAlin (124.11.128.7), 11/29/2014 03:35:58
討論串 (同標題文章)