17. 解：(1)；．

(2)；．

證明：①由已知，得，，．

又，．．

在和中，

，，，

，．

②如圖2，延長(zhǎng)交于點(diǎn)．

，．

在中，，又，

．

．．

(3)成立．

證明：①如圖3，，．

又，．．

在和中，

，，，

．．

②如圖4，延長(zhǎng)交于點(diǎn)，則．

，．

在中，，

．．

．

15. 解：(1) 3-；　　　　　　　　

(2)30°；　　　

　 　(3)證明：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)證明：∵AD∥BC　

　 ∴∠F＝∠DAE

又∵∠FEC＝∠AED

CE＝DE

∴△FEC≌△AED

∴CF＝AD

(2)當(dāng)BC＝6時(shí)，點(diǎn)B在線段AF的垂直平分線上

其理由是：

∵BC＝6 ，AD＝2 ，AB＝8

∴AB＝BC+AD

又∵CF＝AD ，BC+CF＝BF

∴AB＝BF

∴點(diǎn)B在AF的垂直平分線上。

14. 證明：(1)平分，．

在和中，

．

(2)連結(jié)．

，

，

．

，．

．

，．

，．

．

又是公共邊，．

．

13. 證明： 四邊形和四邊形都是正方形

12.證明：(1)在和中

．

(2)，．又，．

11.

解：(1)如圖1；

(2)如圖2；

(3)4．　　　 (8分)

10. 證明：，

，．、)

又，

．

又，

．　　　 (6分

9. 證明： AC∥DE， BC∥EF，又AC=DE, ∴AB=DF　 ∴AF=BD

8. 證明：(1)①

，

·················································································································· 3分

②由得，

分別是的中點(diǎn)，························································· 4分

又

，即為等腰三角形······································································ 6分

(2)(1)中的兩個(gè)結(jié)論仍然成立．············································································· 8分

(3)在圖②中正確畫出線段

由(1)同理可證

又

，和都是頂角相等的等腰三角形······································· 10分

，

　 12分

7. (Ⅰ)證明　 將△沿直線對(duì)折，得△，連，

則△≌△． 　　························································································· 1分

有，，，．

又由，得 ．　 ········································· 2分

由，

，

得． ··································································································· 3分

又，

∴△≌△．　　 ···························································································· 4分

有，．

∴．····························································· 5分

∴在Rt△中，由勾股定理，

得．即． ························································ 6分

(Ⅱ)關(guān)系式仍然成立．　 ····························································· 7分

證明　 將△沿直線對(duì)折，得△，連，

則△≌△． ···················································· 8分

有，，

，．

又由，得 ．

由，

．

得．　 ································································································ 9分

又，

∴△≌△．

有，，，

∴．　

∴在Rt△中，由勾股定理，

得．即．························································ 10分