數學 Plane Geometry mc

2010-09-17 3:29 am

圖片參考：http://imgcld.yimg.com/8/n/HA00543461/o/701009160095413873371580.jpg

1.In the figure 1,BK and CH are the angle bisector of angle ABC and angle
ACD respectively,and they meet at I. Let angle=x, angle BIC=

A. 180-x
B. 90-x
C. 90+x
D. 90+(x/2)

2.In the figure 2,AE,BFand CD are the angle bisector of angle XAB,
angle YBC and angle ZCA respectively.Which of the following must be
true?

A. angle BAC= angle EDF
B. angle ABC= angle EFD
C. angle ACB= angle EDF
D. angle EAB= angle EFD

3.In the figure 3, IB and IC are the bisectors of angle ABC and angle ACB
respectively,angle BIC= 134. angle BAC=

A. 46
B. 56
C. 67
D. 88

4.In the figure 4, BE//CF, CE//DF , AB=3 ,BC=1 ,CD=

A. 1
B. 2/3
C. 4/3
D. 5/2

提供答案就ok
最好有少少解釋

回答 (1)

用戶2053

2010-09-17 9:50 am

✔ 最佳答案

1)ㄥBIC
= 180 - (1/2)(ㄥB + ㄥC)
= 180 - (1/2)(180 - x)
= 180 - 90 + x/2
= 90 + x/2選(D)
2) (Figure 3)設ㄥEAB = ㄥEAX = a
ㄥFBY = ㄥFBC = b
ㄥDCZ = ㄥDCA = c則 △ABC外角和 = 2a + 2b + 2c = 360==> a + b + c = 180
==> a = 180 - b - c而ㄥEFD = ㄥBFC = 180 - b - c = a = ㄥEAB選(D)
3) (Figure 2)2(180 - 134) + ㄥBAC = 180
ㄥBAC = 88
選(D)
4)AE : EF = AB : BC = 3 : 1又AC : CD = AE : EF ==> AC : CD = 3 : 1==> (3+1) : CD = 3 :１==> CD = 4/3選(C)

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