学习必备欢迎下载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：CFACB平分，∴12.⋯⋯⋯⋯⋯⋯⋯⋯1分又 DCAC,∴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,∴2()AEFABDSAESAB.⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯4分又 12AEAB,6AEFABDABDBDFESSSS四边形,⋯⋯⋯⋯⋯⋯5分∴261()2ABDABDSS,⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯6分∴8ABDS,∴ABD的面积为8.⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯7分21FEDCBA学习必备欢迎下载如图，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：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∴△CDA≌△DCE．·······························5分或②△BAD≌△DCE的理由是： AD∥BC，∴∠CDA=∠DCE．···················································································3分FEDCBA（第23题图）FEDCBAG学习必备欢迎下载又 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：ADECBF△≌△．（5分）（2）若AD⊥BD，则四边形BFDE是什么特殊四边形？请证明你的结论...
