E is the point on BC ,
such that AE is the angle bisector of angle CAB
,AE=2CD ,find angle B ,with proof.
更新1:
Thanks 回答者： ☂雨後陽光☀ ( 知識長 )
Maths olympiad (Geometry)
2012-02-03 3:04 am
In isos. triangle ABC,AC=BC,D is the point on AB such that AD=DB,
E is the point on BC ,
such that AE is the angle bisector of angle CAB
,AE=2CD ,find angle B ,with proof.
E is the point on BC ,
such that AE is the angle bisector of angle CAB
,AE=2CD ,find angle B ,with proof.
回答 (1)
2012-02-03 4:49 am
✔ 最佳答案
圖片參考：http://imgcld.yimg.com/8/n/HA04628698/o/701202020069713873434060.jpg
Let DF // AE , then
ㄥEAB = ㄥFDB = ㄥB/2
△AEB ~ △DFB ,
by mid-point theorem , DF = AE/2 = CD
So △CDF is an issos. △ ,
we have
ㄥDCF
= ㄥDFC
= ㄥFDB + ㄥFBD
= ㄥB/2 + ㄥB
By ㄥsum of △ ,
ㄥDCF + ㄥB + ㄥCDB = 180°
(ㄥB/2 + ㄥB) + ㄥB + 90° = 180°
ㄥB = 18°
2012-02-02 20:51:15 補充：
Sorry , it should be ㄥB = 36°.
收錄日期: 2021-04-21 22:26:54
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