内容正文:
19. 证明:∵ AB∥CD∥EF,∴ ∠1 = ∠AEF,
∠3 = 180°-∠CEF.
又∵ ∠2 = ∠AEF-∠CEF,
∴ ∠3+ ∠1 - ∠2 = ( 180° - ∠CEF) + ∠AEF -
( ∠AEF - ∠CEF) = 180° - ∠CEF + ∠AEF -
∠AEF+∠CEF= 180°.
20. 证明:∵ CE 平分∠DCF,
∴ ∠2 = ∠DCE.
又∵ ∠1 = ∠2,
∴ ∠1 = ∠DCE,
∴ AB∥CD,
∴ ∠FHB= ∠DCF.
21. 解:( 1) ∵ BE 平分∠ABC, ∴ ∠ABC = 2 ∠ABE. ∵
∠ABC= 2∠E,∴ ∠ABE= ∠E,∴ AB∥EF;
(2)AF⊥BE,理由如下:∵ ∠ADE = ∠BCD. ∴ AD
∥BC,∴ ∠DAB+ ∠CBA = 180°. ∵ BE 平分
∠ABC,AF 平分∠BAD,∴ ∠ABC = 2∠ABE,
∠BAD = 2∠BAF. ∵ 2 ∠ABE + 2 ∠BAF =
180°,∴ ∠ABE + ∠BAF = 90°, ∴ ∠AOB =
180°-( ∠ABE+∠BAF) = 180° - 90° = 90°,
∴ AF⊥BE.
22. 解:(1)直线 AD 与 BC 互相平行,理由:∵ AB∥CD,
∴ ∠A + ∠ADC = 180°, 又 ∵ ∠A = ∠C, ∴
∠ADC+∠C= 180°,∴ AD∥BC;
(2) ∵ AB∥CD,∴ ∠ABC = 180° - ∠C = 80°, ∵
∠DBF= ∠ABD,BE 平分∠CBF,∴ ∠DBE =
1
2
∠ABF+ 1
2
∠CBF= 1
2
∠ABC= 40°;
(3) 存在. ∵ AB∥CD, ∴ ∠BDC = ∠ABD, 设
∠ABD = ∠DBF = ∠BDC = x°. ∴ ∠BEC =
∠ABE = x° + 40°, ∴ ∠ADC = 180° - ∠A =
80°,∴ ∠ADB = 80°-x°. 若∠BEC = ∠ADB,
则 x° + 40° = 80° - x°,得 x° = 20°. ∴ 存在
∠BEC= ∠ADB,∠ADB= 60°.
23. 解:( 1) 44° 【解析】 ∵ AB∥CD,∠MAB = 46°,∴
∠ACE= ∠MAB= 46°,∵ ME⊥AC,∴ ∠CME
= 90°,∴ ∠MEC= 90°-∠MCE= 44°;
(2) ∠MAB = 90° + ∠MEC,理由如下: ∵ ME⊥
AC, ∴ ∠CME = 90°, ∴ ∠MCE = 90° -
∠MEC,∵ AB∥CD,∴ ∠BAC = ∠MCE = 90°
- ∠MEC, ∵ ∠MAB = 180° - ∠BAC, ∴
∠MAB = 180° - ( 90° - ∠MEC ) = 90° +
∠MEC;
(3)∠MEC+∠MAB = 90°,理由如下:过点 M 作
EM⊥AC 交 DC 的延长线于点 E. ∵ AB∥CD,
∴ ∠MAB = ∠MCE,∵ ME⊥AC,∴ ∠CME =
90°,∴ ∠MEC = 90°-∠MCE = 90°-∠MAB,
∴ ∠MEC+∠MAB= 90°.
第六章追梦过关检测卷
一、1. B 【解析】27 的相反数为- 27,则 3 -27 = - 3. 故选
B.
2. C 3. D
4. A 【解析】B. 4 = 2,错误;C. (-5) 2 = 5,错误,D.
3 -27 = -3,错误. 故选 A.
5. B 6. B
7. A 【解析】A. - | -2 | = -2, 3 8 = 2,- 2 与 2 互为相反
数,正 确; B. - (-3) 2 = - 3, - 3 = - 3, 错 误; C.
| 3 -3 | = 3 3,3 3 = 3 3,错误;D. - 5 ≠
1
5
,错误. 故选
A.
8. C
9. D 【解析】∵ | a | = 5, b = 6,∴ a = ± 5,b = 6. 当 a =
5,b= 6 时,2a-b= 4;当 a= -5,b = 6 时,2a-b = -16,∴
2a-b 的值是 4 或-16. 故选 D.
10. A 【解析】若 n= 10,∵ 10b = 101,∴ b = 1,根据题意
可得 d(10)= 1,则 d(0. 8) = d( 2
10
× 4) = d( 2
10
) +
d(4)= d(2)-d(10)+d(2×2)= d(2)-d(10)+d(2)+
d(2)= 3d(2)-d(10)= 3×0. 301-1= -0. 097. 故选 A.
二、11. 7 ,π, 3 36
12. -4 【解析】由题意可得 m+2 = 0,n-