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，

．

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

又

，和都是顶角相等的等腰三角形······································· 10分

，

　 12分

7. (Ⅰ)证明　 将△沿直线对折，得△，连，

则△≌△． 　　························································································· 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)关系式仍然成立．　 ····························································· 7分

证明　 将△沿直线对折，得△，连，

则△≌△． ···················································· 8分

有，，

，．

又由，得 ．

由，

．

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

又，

∴△≌△．

有，，，

∴．　

∴在Rt△中，由勾股定理，

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

6. 解：(1)证明：，，，

，

，

．

(2)①是；②是；③否．

②的证明：如图，

，，，

，

，

，

．

③的证明：如图，

，，

，

．又，

，

，即．

5. 解:⑴证明:∵AC平分∠MAN，∠MAN＝120°，

∴∠CAB＝∠CAD＝60°，

∵∠ABC＝∠ADC＝90°，

∴∠ACB＝∠ACD＝30°，…………1分

∴AB＝AD＝AC，……………………2分

∴AB+AD＝AC。……………………3分

⑵成立。……………………………r…4分

证法一：如图，过点C分别作AM、AN的垂线，垂足分别为E、F。

∵AC平分∠MAN，∴CE＝CF.

∵∠ABC+∠ADC＝180°,∠ADC+∠CDE＝180°,

∴∠CDE＝∠ABC,………………………………………………………………5分

∵∠CED＝∠CFB＝90°，∴△CED≌△CFB,∴ED＝FB,……………………6分

∴AB+AD＝AF+BF+AE－ED＝AF+AE,由⑴知AF+AE＝AC,

∴AB+AD＝AC……………………………………………………………………7分

证法二：如图，在AN上截取AG＝AC，连接CG.

∵∠CAB＝60°,AG＝AC,∴∠AGC＝60°,CG＝AC＝AG,…………5分

∵∠ABC+∠ADC＝180°,∠ABC+∠CBG＝180°,

∴∠CBG＝∠ADC,∴△CBG≌△CDA,……………………………………6分

∴BG＝AD,

∴AB+AD＝AB+BG＝AG＝AC，…………………………………………7分

⑶①;………………………………………………………………………8分

②.………………………………………………………………………9分

证明：由⑵知，ED＝BF,AE＝AF，

在Rt△AFC中，,即,

∴,………………………………………………………………10分

∴AB+AD＝AF+BF+AE－ED＝AF+AE＝2，…………11分

4. 证明：(1)证明：方法一：在△ACD和△BCE中，

AC＝BC，

∠DCA＝∠ECB＝90°，

DC＝EC，　

∴ △ACD≌△BCE(SAS)． ………………2分

∴ ∠DAC＝∠EBC．　 ………………………3分

　　 ∵ ∠ADC＝∠BDF，　

　　 ∴ ∠EBC+∠BDF＝∠DAC+∠ADC=90°．

　　 ∴ ∠BFD=90°．　

∴ AF⊥BE．　 …………………………………5分　

方法二：∵ AC＝BC，DC＝EC，

∴ ．即tan∠DAC＝tan∠EBC．　

∴ ∠DAC＝∠EBC．(下略)…………………3分

(2)AF⊥BE．　 …………………………………6分　

∵ ∠ABC＝∠DEC＝30°，∠ACB＝∠DCE＝90°，

∴ ＝tan60°．　 ……………………7分

∴ △DCA∽△ECB． 　…………………………8分

∴ ∠DAC＝∠EBC．　 …………………………9分

∵ ∠ADC＝∠BDF，

∴ ∠EBC+∠BDF＝∠DAC+∠ADC=90°．　

∴ ∠BFD=90°．　

∴ AF⊥BE．　　 ……………………………………………………………………10分