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
3. (1)证明:在ΔABC和ΔDCB中
∴ΔABC≌ΔDCB(SSS)
(2)等腰三角形。
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