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12.证明:(1)在

(2).又

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11.

解:(1)如图1;

(2)如图2;

(3)4.    (8分)

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10. 证明:

.、)

.    (6分

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

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8. 证明:(1)①

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

②由

分别是的中点,························································· 4分

,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

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

  12分

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7. (Ⅰ)证明  将△沿直线对折,得△,连

则△≌△.   ························································································· 1分

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

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

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

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

∴在Rt△中,由勾股定理,

.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

又由,得

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

∴△≌△

. 

∴在Rt△中,由勾股定理,

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

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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

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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

.………………………………………………………………………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

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4. 证明:(1)证明:方法一:在△ACD和△BCE中,

ACBC

DCA=∠ECB=90°,

DCEC, 

∴ △ACD≌△BCE(SAS). ………………2分

∴ ∠DAC=∠EBC.  ………………………3分

   ∵ ∠ADC=∠BDF, 

   ∴ ∠EBC+∠BDF=∠DAC+∠ADC=90°.

   ∴ ∠BFD=90°. 

AFBE.  …………………………………5分 

方法二:∵ ACBCDCEC

.即tan∠DAC=tan∠EBC. 

∴ ∠DAC=∠EBC.(下略)…………………3分

(2)AFBE.  …………………………………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°. 

AFBE.   ……………………………………………………………………10分

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3. (1)证明:在ΔABC和ΔDCB中

∴ΔABC≌ΔDCB(SSS)

(2)等腰三角形。

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