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2013年中考数学知识点:四边形——四边形专题复习

发布时间:2014-03-17 17:37:35  

三角形与四边形

如图5,在△ABC中,BC>AC, 点D在BC上,且DC=AC,∠ACB的平分线CF交AD于F,点

E是AB的中点,连结EF.

(1)求证:EF∥BC.

(2)若四边形BDFE的面积为6,求△ABD的面积.

(1)证明:

? CF平分?ACB,

∴ ?1??2.……………………1分

又∵ DC?AC,

∴ CF是△ACD的中线,

∴ 点F是AD的中点.…………2分

∵ 点E是AB的中点,

∴ EF∥BD,

即 EF∥BC. …………………………3分

(2)解:由(1)知,EF∥BD,

∴ △AEF∽△ABD ,

∴ B2AEF1DCS?AEFAE2?().……………………………………4分 S?ABDAB

又∵ AE?1AB, 2

S?AEF?S?ABD?S四边形BDFE?S?ABD?6,………………5分

∴ S?ABD?612?() ,………………………………………6分 S?ABD2

∴ S?ABD?8,

∴ ?ABD的面积为8. ………………………………………7分

如图,四边形ABCD是菱形,DE⊥AB交BA的延长线于E,DF⊥BC,交BC的延长

线于F。

请你猜想DE与DF的大小有什么关系?并证明你的猜想。

解:DE=DF

证明如下:

连结BD

∵四边形ABCD是菱形

∴∠CBD=∠ABD(菱形的对角线平分一组对角)

∵DF⊥BC,DE⊥AB

∴DF=DE(角平分线上的点到角两边的距离相等)

如图,在等腰梯形ABCD中,已知AD∥BC,AB=DC,AD=2,BC=4,延长BC到E,使

CE=AD.

(1)写出图中所有与△DCE全等的三角形,并选择其中一对说明全等的理由;(5分)

(2)探究当等腰梯形ABCD的高DF是多少时,对角线AC与BD互相垂直?请回答并说明

理由.(5分)

解:

F C (第23题图) E

解:(1)△CDA≌△DCE,△BAD≌△DCE; ······························································· 2分

① △CDA≌△DCE的理由是:

∵AD∥BC,

∴∠CDA=∠DCE. ····························· 3分

又∵DA=CE,CD=DC , ······························· 4分

F C E

∴△CDA≌△DCE. ······································· 5分

或 ② △BAD≌△DCE的理由是:

∵AD∥BC,

∴∠CDA=∠DCE. ········································································································· 3分 又∵四边形ABCD是等腰梯形,

∴∠BAD=∠CDA,

∴∠BAD =∠DCE. ········································································································ 4分 又∵AB=CD,AD=CE,

∴△BAD≌△DCE. ··································································································· 5分

(2)当等腰梯形ABCD的高DF=3时,对角线AC与BD互相垂直. ··················· 6分 理由是:设AC与BD的交点为点G,∵四边形ABCD是等腰梯形,

∴AC=DB.

又∵AD=CE,AD∥BC,

∴四边形ACED是平行四边形, ············································································ 7分 ∴AC=DE,AC∥DE.

∴DB=DE. ············································································································ 8分 则BF=FE,

又∵BE=BC+CE=BC+AD=4+2=6,

∴BF=FE=3. ········································································································· 9分 ∵DF=3,

∴∠BDF=∠DBF=45°,∠EDF=∠DEF=45°,

∴∠BDE=∠BDF+∠EDF=90°,

又∵AC∥DE

∴∠BGC=∠BDE=90°,即AC⊥BD. ·································································· 10分 (说明:由DF=BF=FE得∠BDE=90°,同样给满分.)

如图8

,在ABCD中,E,F分别为边AB,CD的

中点,连接E、BF、BD.

(1)求证:△ADE≌△CBF.(5分)

C

A E

(图8) B

(2)若AD⊥BD,则四边形BFDE是什么特殊

四边形?请证明你的结论.(5分)

(1)在平行四边形ABCD中,∠A=∠C,AD=CD,

∵E、F分别为AB、CD的中点

∴AE=CF ……………………………………………………2分

?AD?CB?在?AED和?CFB中,??A??C (2)?AE?CF?

??AED??CFB?SAS?.........................................................................5 分

若AD⊥BD,则四边形BFDE是菱形. …………………………1分 证明:?AD?BD,

??ABD是Rt?,且AB是斜边(或?ADB?90o).....................2 分

?E是AB的中点,

1 ?DE?AB?BE ......................................................................3分2

由题意可知EB//DF且EB?DF,

?四边形BFDE是平行四边形

? 四边形BFDE是菱形. .............................................................5分

如图,在□ABCD中,BC=2AB=4,点E、F分别是BC、AD的中点.

(1)求证:△ABE≌△CDF;

(2)当四边形AECF为菱形时,求出该菱形的面积.

(1)证明略; ····················································································································· (4分)

(2)当四边形AECF为菱形时,△ABE为等边三角形, ·············································· (6分) 四边形ABCD的高为, ···························································································· (7分)

E C

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