1 trigonometry Q.
2011-08-12 10:39 pm
圖片參考：http://imgcld.yimg.com/8/n/HA00936359/o/701108120063213873441740.jpg
For any ∆ABC (Beside isos. ∆),
AJ is an angle bisector of ∆BAC, so ∠BAJ=∠CAJ.
HK is a perpendicular bisector of BC, so HK⊥BC,BK=CK.
Extend AJ and HK, they meet at point O.
Prove that point O is outside the ∆ABC (Beside isos. ∆).
回答 (1)
2011-08-13 6:56 am
✔ 最佳答案
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圖片參考：http://tw.blog.yahoo.com/photo/photo.php?id=5cMJAGuFERmdAf4sfsa5M35I&photo=ap_F23_20110812105341246.jpg
http://tw.myblog.yahoo.com/sincos-heart/article?mid=3406&prev=1185&next=3402
2011-08-13 00:02:51 補充：
更正：做BC中點K 連接 OK
2011-08-13 16:04:23 補充：
how to prove when ∠BAO=∠CAO,arc OB will =arc OC??
兩個圓周角相同，必對應於兩個相同的弧。
OB弧 對應於 ∠BAO，OC弧 對應於 ∠CAO
2011-08-13 16:18:59 補充：
how to prove when ∠BAO=∠CAO,arc OB will =arc OC??
證明：
http://tw.myblog.yahoo.com/sincos-heart/article?mid=3406
2011-08-14 01:10:27 補充：
why (∠3)/2 =∠1?
圓周角 是 圓心角的一半
2011-08-14 01:27:45 補充：
證明：
http://tw.myblog.yahoo.com/sincos-heart/article?mid=3406
(在最底下)
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110812000051KK00632