1.http://img7.photo.qooza.hk/userphoto/7/8/178978/292882/12386843fd11422a.jpg
圖中所示為四邊形ABCD,BE是角ABC的角平分線.若角BEC-90度及BA平行於CD,證明EC是角BCD的角平分線.
2.http://img7.photo.qooza.hk/userphoto/7/8/178978/292882/12386844fa044762.jpg
如圖所示,在三角形ABC中,AB與AC上的垂直平分線相交於D.若D是BC上的一點,證明角A=90度.
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2008-10-18 6:29 pm
回答 (3)
2008-10-18 8:42 pm
✔ 最佳答案
本證明題的答案如下：圖片參考：http://i277.photobucket.com/albums/kk75/uncle_michael/20081017_mat_2.jpg?t=1224304834
2008-10-18 7:26 pm
1)因為ABCD是四邊形,所以AD//BC,AB//CD
EBC=BEA(alt.angles,AD//BC)
CED=ECB(alt.angles,AD//BC)
ABC=CDA(opposite.angles equal)
EBC=180-90-EBC(angles sum of triangle)
=90-EBC
ECD=180-(90-EBC)-2EBC
=90-EBC
因為ECD=ECB,所以EC是角BCD的角平分線
2)唔識...
EBC=BEA(alt.angles,AD//BC)
CED=ECB(alt.angles,AD//BC)
ABC=CDA(opposite.angles equal)
EBC=180-90-EBC(angles sum of triangle)
=90-EBC
ECD=180-(90-EBC)-2EBC
=90-EBC
因為ECD=ECB,所以EC是角BCD的角平分線
2)唔識...
參考: 自己~~
2008-10-18 6:58 pm
Q1.
Let ∠ABE and ∠ EBC be x & y respectively.
∠ ECB=180°-90°-y (∠ sum of △)
∠ ECB=90°-y
∠ABC+∠BCD=180° (int.∠s, AB//CD)
∴x+y+(90°-y)+∠ECD=180°
∴2y+90°-y+∠ECD=180°
∴y+∠ECD=90°
∴∠ECD=90°-y
i.e.∠ECD=∠ECB
i.e.CD is the angle bisector of ∠BCD.
Let ∠ABE and ∠ EBC be x & y respectively.
∠ ECB=180°-90°-y (∠ sum of △)
∠ ECB=90°-y
∠ABC+∠BCD=180° (int.∠s, AB//CD)
∴x+y+(90°-y)+∠ECD=180°
∴2y+90°-y+∠ECD=180°
∴y+∠ECD=90°
∴∠ECD=90°-y
i.e.∠ECD=∠ECB
i.e.CD is the angle bisector of ∠BCD.
參考: own knowledge
收錄日期: 2021-04-15 01:06:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081018000051KK00506