18. 证明:
17. 解:(1);.
(2);.
证明:①由已知,得,,.
又,..
在和中,
,,,
,.
②如图2,延长交于点.
在中,,又,
.
..
(3)成立.
证明:①如图3,,.
②如图4,延长交于点,则.
在中,,
15. 解:(1) 3-;
(2)30°;
(3)证明:在△AEF和△D′BF中,
∵AE=AC-EC, D’ B=D’ C-BC,
又AC=D’ C,EC=BC,∴AE=D’ B.
又 ∠AEF=∠D’ BF=180°-60°=120°,∠A=∠CD’E=30°,
∴△AEF≌△D’ BF.∴AF=FD’
16. (1)证明:∵AD∥BC
∴∠F=∠DAE
又∵∠FEC=∠AED
CE=DE
∴△FEC≌△AED
∴CF=AD
(2)当BC=6时,点B在线段AF的垂直平分线上
其理由是:
∵BC=6 ,AD=2 ,AB=8
∴AB=BC+AD
又∵CF=AD ,BC+CF=BF
∴AB=BF
∴点B在AF的垂直平分线上。
14. 证明:(1)平分,.
(2)连结.
,
又是公共边,.
13. 证明: 四边形和四边形都是正方形
12.证明:(1)在和中
(2),.又,.
11.
解:(1)如图1;
(2)如图2;
(3)4. (8分)
10. 证明:,
,.、)
又,
. (6分
9. 证明: AC∥DE, BC∥EF,又AC=DE, ∴AB=DF ∴AF=BD
8. 证明:(1)①
·················································································································· 3分
②由得,
分别是的中点,························································· 4分
又
,即为等腰三角形······································································ 6分
(2)(1)中的两个结论仍然成立.············································································· 8分
(3)在图②中正确画出线段
由(1)同理可证
,和都是顶角相等的等腰三角形······································· 10分
12分