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# 2013年中考数学知识点：四边形——四边形专题复习

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

∴ CF是△ACD的中线，

∵ 点E是AB的中点，

∴ EF∥BD,

（2）解：由（1）知，EF∥BD，

∴ △AEF∽△ABD ,

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

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

∵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)

（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

F C （第23题图） E

① △CDA≌△DCE的理由是：

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

F C E

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

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

（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．

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

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

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

∴∠BDF=∠DBF=45°，∠EDF=∠DEF=45°，

∴∠BDE=∠BDF+∠EDF=90°，

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

，在ABCD中，E，F分别为边AB，CD的

C

A E

（图8） B

∵E、F分别为AB、CD的中点

∴AE=CF ……………………………………………………2分

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

?E是AB的中点,

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

?四边形BFDE是平行四边形

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

（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