求證：任何一個三角形的三條角平分線相交於一點。
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數學知識交流---求證
2011-10-12 4:17 am
回答 (2)
2011-10-12 5:12 am
✔ 最佳答案
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ㄥB 平分線 BE 及 ㄥC 平分線 CF 必交於 O ,由 O 引 AB , BC , CA 垂線 Of , Od , Oe ,則
OB = OB (公共)
ㄥOBf = ㄥOBd (已知)
ㄥOfB = ㄥOdB = 90° (垂線)
故 △OfB ≡ △OdB (SAA)
得 Of = Od 同理 , △OdC ≡ △OeC
得 Od = Oe 綜合得 Of = Oe
而 OA = OA (公共)
ㄥAfO = ㄥAeO = 90° (垂線)
∴
△AfO ≡ △AeO (RHS)故ㄥOAB = ㄥOAC ,
AO 亦為 ㄥA 之平分線 , 故證。
2011-10-12 4:50 am
We shall use two theorems.
Theorem 1: Ceva's
For triangle ABC, line AD, BE CF (D,E,F are on BC, AC and AB respectively) intersect at one point if and only if
AF/FB * BD/DC * CE/EA = 1
Theorem 2: Angle bisector theorem.
For triangle ABC, AD is the angle bisector of angle BAC where D is on BC.
Then AB:AC = BD : CD.
Considering the angle bisectors of the triangle,
AF/FB * BD/DC * CE/EA
= AC/BC * AB/AC * BC/AB = 1
Therefore angle bisectors of a triangle are concurrent.
Theorem 1: Ceva's
For triangle ABC, line AD, BE CF (D,E,F are on BC, AC and AB respectively) intersect at one point if and only if
AF/FB * BD/DC * CE/EA = 1
Theorem 2: Angle bisector theorem.
For triangle ABC, AD is the angle bisector of angle BAC where D is on BC.
Then AB:AC = BD : CD.
Considering the angle bisectors of the triangle,
AF/FB * BD/DC * CE/EA
= AC/BC * AB/AC * BC/AB = 1
Therefore angle bisectors of a triangle are concurrent.
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