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1.
在△ ABC中 ,∠ C=90°,CD⊥ AB于 D,AM是∠ BAC的平分线 ,交 CD于 E,交 BC于 M,过E作 EF∥ AB交 BC于 F。求证 :CM=BF。证法一 :(运用三角形知识 )证明 :过 M作 MN⊥ AB于点 N。∵∠ 1=∠ 2 ,易证△ ACM≌△ ANM,∴CM=MN。  ( 1)又 CD⊥ ABMN⊥ AB CD∥ MN, ∠ 3=∠ 5∠ 4 =∠ 5 ∠ 3=∠ 4 CE=CM。  ( 2 )由 ( 1)、( 2 )得 CE=MN。在 Rt△ EFC和 Rt△ NBM中 ,EF∥ AB ∠ B=∠ CFE,∠ CEF=∠ MNB,CE=MN Rt△ EFC≌ Rt△ NBM,∴ CF=BM,∴ CM=BF。  证法二 :(运用四边形知识 )证明 :过 M…  相似文献   

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李成章 《中等数学》2005,(11):F0004-F0004
题目分别以△ABC的边AB、AC为一边向形外作△ABF和△ACE,使得△ABF∽△ACE,且∠ABF=90°.求证:BE、CF和边BC上的高AH三线共点.分析:因为AH为边BC上的高,故可想到构造一个三角形,使得所证的三条线恰为这个三角形的三条高所在的三条直线.当然图1交于一点.证明:如图1,过点B作BD⊥CF于点D,延长BD、HA交于点M,过点C作CG⊥BE于点G,延长CG、HA交于点M′.于是,只须证明M′与M重合.因为MH⊥BC,MB⊥CF,所以,∠DCB=∠BMH.又∠ABF=90°=∠BDF,因此,∠MBA=∠BFD.故△MBA∽△CFB.则BMCA=FABB,MA=BCF.BAB.同理,…  相似文献   

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题目1:已知,如图1,在矩形 ABCD 中,点E,F 分别在 BC、CD 上,且 CE=AB,CF=BE求证:AE⊥EF.证明:由条件可得△ABE≌△ECF,所以∠1=∠2,又∠B ∠1 ∠3=180°,∠AEF ∠3 ∠2=180°,所以∠AEF=∠B=∠C=90°,所以 AE⊥EF.  相似文献   

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1.利用线段之间的关系例1 如图l,已知GD⊥CB,AC交GD于F,AE⊥CB,又∠G=∠GFA. 求证:AE平分∠CAB. 证明由GD⊥CB,AE⊥CB,得GD∥AE.则∠CAE=∠GFA=∠G.由∠FAG=180°-2∠G,知图1  相似文献   

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在证明三角形全等时,有些同学常出现种种错误.下面举例说明,以引起注意.例1已知:如图1,AC⊥BC,DC⊥EC,AC=BC,DC=EC,求证:∠D=∠E.错证:在△ACE与△CBD中,∵AC⊥BC,DC⊥EC,∴∠ACB=∠ECD=90°,AC=BC,DC=EC.∴△ACE≌△CBD.∴∠D=∠E.评析:上面的证明中,错误地应用了“SAS”,但∠ACB与∠ECD并不是这一对三角形中的内角.也就不是AC与CE、BC与CD的夹角,错误原因是未能深刻理解“SAS”判定方法.!正确证明:∵AC⊥BC,DC⊥EC,∴∠ACB=∠ECD=90°.∴∠ACE=∠BCD.在△ACE与△CBD中,∵AC=BC,∠ACE=∠BCD,…  相似文献   

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定理 1:若△DEF是△ABC的垂足三角形,则△DEF的三边长分别为acosA、bcosB、CcosC.(如图1) 证明:因为BE⊥AC,CF⊥AB,所以∠BEC=∠CFB=90°,所以B、C、E、F四点共圆.所以∠AEF=∠ABC,又因为∠EAF=∠BAC.所以B△AEF∽△ABC,所以EF/BC=AE/AB,在Rt△ABE中,cosA=AE/AB,所以EF/BC=cosA,所以,EF=acosA,同理可得DF=bcosB,DE=ccosC  相似文献   

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1.巧构全等三角形证线段相等例 1.已知 ,如图 ,AB=DE,直线 AE、BD相关于点 O,∠ B与∠ D互补。  求证 :AO=ED。证明 :过点 A作 AC∥ DE交 BD于 C,则∠ D=∠ 2。∵∠ 1 ∠ 2 =180°,∠ B ∠ D=180°,∴∠ 1=∠ B,∴ AB=AC,∴ AB=DE=CA。在△ ACO和△ EDO中 ,∠ AOC=∠ EOD,∠ 2=∠ D,AC=DE;∴△ ACO △ EDO( AAS) ,∴ AO=ED。2 .巧构全等三角形证角相等例 2 .已知等边△ ABC的边长为 a,在 BC的延长线上取一点 D,使 CD=b,在 BA延长线上取一点 E,使 AE=a b。求证 :∠ ECD=∠ EDC。证明 :过 E作 EF∥ AC…  相似文献   

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利用三角形全等可证明线段相等,以及证明与线段相等有关的线段和、差、倍、分等问题;还可证明两角相等,以及证明与两角相等有关的线段平行、线段垂直等问题.例1如图,∠BAC=90°,AB=AC,F是BC上一点,BD⊥AF于D,E为AF延长线上一点,CE⊥AE,求证:DE=AE-CE.证明:∵CE⊥AE,BD⊥AF于D,∴∠AEC=∠BDA=90°.∴∠1=90°-∠3=∠2.在△AEC和△BDA中,∵∠1=∠2,∠AEC=∠BDA,AC=AB,∴△AEC≌△BDA.∴CE=AD.∵DE=AE-AD,∴DE=AE-CE.例2如图,在△ABC中,D是AB的中点,DE∥BC交AC于E,F是BC上的点,BF=DE,求证:DF∥AC.证…  相似文献   

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在一次统一招生数学试题中有这样一道题;在四边形ABCD中,AB=CD,E、F为AD、BC的中点(如图1),延长EF交BA的延长成于G,交CD的延长线于H,求证∠BGF=∠CHF。本题证法很多,其中有一位考生是这样证明的: 连结EC,将△DCE绕E点顺时针方向旋转180°至△AC’E,D点转到A点的位置,C点转到C’的位置,这时,若连结C’B,则有∠3=∠4,故要证∠1=∠2,只要证明C’B∥EF,而已知BF=FC,又CF=FC’,得EF是△CC’B的中位线,问题得以解决。  相似文献   

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一题多证(解),是培养思维灵活性的好方法.扬州市曾有一道数学中考题,用它来作一题多证的练习很有益.已知:正方形ABCD的对角线相交于O,EF∥AB,并分别交OA,OB于E,F(图1).求证:(1)BE=CF;(2)BE⊥CF.图1不难看出,本题融合了三角形和四边形的不少知识,证明思路是比较宽广的.本题结论的第一部分是求证两线段相等,我们很自然地想到先找出这两条线段分别所在的三角形,设法证明这两个三角形全等.证明问题的第二部分时,最好能用上第一部分的证明.证法一(见图2)图2(1)∵正方形ABCD的对角线相交于O,∴BO=CO,且∠BOE=∠COF=90°.又EF∥AB…  相似文献   

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This article indicates how the application of a simplified version of the analytical abstraction method (AAM) was used in curriculum development for consumer learning at one higher education institution in South Africa. We used a case study design and qualitative research methodology to generate data through semi‐structured interviews with eight learning facilitators at the Cape Peninsula University of Technology. This data set forms the basis of the reported research. Application of basic‐ and higher‐level analysis resulted in the identification of patterns that confirmed the need for consumer learning and informed the situation analysis with regard to a ‘readiness climate’ at the institution. We also gained insight into aspects that need to be considered during curriculum development for consumer learning as the AAM has proved to be a useful guiding tool in developing a structured explanatory framework for curriculum development. The article concludes with the view that the promotion of consumer learning in university curricula has been under‐researched and that, despite current efforts, university curricula are slow to adopt consumer learning as a critical learning outcome. We suggest several possibilities that might assist in overcoming this inertia.  相似文献   

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<正>Simone de Beauvoir said,"A woman is not born but made."Then,what makes women?Culture?Parental teaching?Social environments?...When women are born,they are told they are a girl and brought up in the way that girls grow up.Gradually,an invisible and powerful voice is rooted in their mind,"I’m a girl and that I’ll be a woman or housewife."  相似文献   

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Reflecting from the Philippine experience, this article explores an emerging picture that characterizes contemporary Bible reading attitudes of college students. Six new attitude factor definitions are developed following the development of the Bible Reading (BR) attitude scale for college students constructed by this author in a separate study. Through empirical procedures, the pre-tested scale produced six new attitude factor definitions that displaced the nine original pre-test attitude sub-factors. Their Bible reading attitudes are described in terms of motivation, reading preference, and acquaintance. This research provides practitioners with useful attitudinal dimensions for youth formation particularly in biblical formation and religious education.  相似文献   

15.
人草大战     
潘晨曦 《海外英语》2006,(11):32-33
我突然怀念起曾经住过的在城市陋屋区的旧房子来,陋室在12层的公寓上,没有电梯,只有一间屋子,半个卫生间。事情是这样的:这一切发生得非常简单。我们搬到乡下后不久,我的妻子,詹妮弗,觉得后院需要一次彻底的整理。“后院怎么了?”我问道“,看看我们院子里的草长得多壮实,多招摇。我敢打赌,这一带可没有哪家能比得上我们。”这时候詹妮弗透露了一个小秘密,我们草坪上的根本就不是草,只有壮实、招摇的毒葛。我认为,毒葛长得欣欣向荣,可不像后院的其他植物,我们也许应该除掉其他东西,比如说那个看起来病怏怏的玫瑰。我据理力争“:为什么要把真…  相似文献   

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King for a Day     
It was breakfast time. "Presto told me that it's an old folk custom to elect (选举) a king around this time of year," Buzz said excitedly.* "And the others have to* grant(准予)the king all his wishes.*"  相似文献   

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