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17. 解:(1)

(2)

证明:①由已知,得

中,

②如图2,延长于点

中,,又

(3)成立.

证明:①如图3,

中,

②如图4,延长于点,则

中,

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15. 解:(1) 3-;        

(2)30°;   

   (3)证明:在△AEF和△DBF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’ BFAF=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。

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14. 证明:(1)平分

中,

(2)连结

是公共边,

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13. 证明: 四边形和四边形都是正方形

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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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