八年级数学下册-小专题四-特殊平行四边形的性质与判定测试题-(新版)湘教版(DOC 5页).doc

1、leadership intention and meaning into the text. Although his eyes now write only a relatively small number of materials, but also quite difficult, even to a molding material herself seems not very satisfied with the quality of service needs to be enhanced.Three is a lot of work to implement passive,

2、 active and advanced. Most contractors do matters assigned by the leadership, work is often in the orders, to implement the task, do not want to lead, not want to think ahead, plan ahead, and to seek the leadership; seek deep into the research, provide information good staff seek. Especially in the

3、writing process of material, it encountered much, time tight, heavy特殊平行四边形的性质和判定1如图，O是矩形ABCD的对角线AC的中点，M是AD的中点，若AB5，AD12，则四边形ABOM的周长为（ ）A17 B18 C19 D202(烟台中考)如图，在菱形ABCD中，M、N分别在AB，CD上，且AMCN，MN与AC交于点O，连接BO，若DAC28，则OBC的度数为（ ）A28 B52 C62 D723(吉林中考)如图，四边形ABCD、AEFG是正方形，点E，G分别在AB，AD上，连接FC，过点E作EHFC，交BC于点H.若A

4、B4，AE1，则BH的长为（ ）A1 B2 C3 D34(玉林、防城港中考)下列命题是假命题的是（ ）A四个角相等的四边形是矩形B对角线相等的平行四边形是矩形C对角线垂直的四边形是菱形D对角线垂直的平行四边形是菱形5(曲靖中考)如图，在矩形ABCD中，E，F分别是AD，BC的中点，连接AF，BE，CE，DF，分别交于点M，N，四边形EMFN是（ ）A正方形 B菱形C矩形 D无法确定6(内江中考)如图，正方形ABCD的面积为12，ABE是等边三角形，点E在正方形ABCD内，在对角线AC上有一点P，使PDPE最小，则这个最小值为（ ）A.B2C2D.7(黔西南中考)如图，将矩形纸片ABCD折叠，使

5、边AB、CB均落在对角线BD上，折痕为BE，则EBF的大小为_8(沈阳中考)如图，在矩形ABCD中，对角线AC，BD相交于点O，点E，F分别在边AD，BC上，且DECF，连接OE，OF.求证：OEOF.9如图，已知两个菱形ABCD，CEFG共顶点C，且点A，C，F在同一直线上，连接BE，DG.(1)在不添加辅助线时，写出其中的两对全等三角形；(2)证明：BEDG.10如图，在矩形ABCD中，对角线BD的垂直平分线MN与AD相交于点M，与BD相交于点O，与BC相交于点N，连接BM，DN.(1)求证：四边形BMDN为菱形；(2)若AB4，AD8，求MD的长11如图，在ABC中，D是BC边的中点，E

6、、F分别在线段AD及其延长线上，CEBF.(1)求证：BDFCDE；(2)若BDDF，求证：四边形BFCE是矩形12已知正方形ABCD的对角线AC与BD交于点O，点E，F分别是OB，OC上的动点(1)如果动点E，F满足BECF(如图1)；写出所有以点E或F为顶点的全等三角形(不得添加辅助线)；证明：AEBF；(2)如果动点E，F满足BEOF(如图2)，问AEBF时，点E在什么位置，并证明你的结论参考答案1D2.C3.C4.C5.B6.B7.458.证明：四边形ABCD为矩形，ADCBCD90，ACBD，ODBD，OCAC.ODOC.ODCOCD.ADCODCBCDOCD，即EDOFCO.又DE

7、CF，ODEOCF.OEOF.9.(1)ADCABC，GFCEFC，GDCEBC(任意两对均可)(2)证明：四边形ABCD，四边形CEFG是菱形，DCBC，CGCE，DCABCA，GCFECF.DCG180DCAGCF，BCE180BCAECF，DCGBCE.GDCEBC(SAS)BEDG.10.(1)证明：MN是BD的垂直平分线，MBMD，OBOD，BONDOM.四边形ABCD是矩形，ADBC.OBNODM.BONDOM.BNMD.四边形BMDN是平行四边形又BDMN，BMDN是菱形(2)设MDx，则AM8x，BMx.在RtABM中，BM2AB2AM2，x242(8x)2.解得x5.即MD5

8、.11.证明：(1)D是BC边的中点，BDDC.CEBF，ECDFBD.在BDF和CDE中，BDFCDE(ASA)(2)BDFCDE，EDDF.又BDCD，四边形EBFC是平行四边形BDDF，BCEF.四边形EBFC是矩形12.(1)ABEBCF，AOEBOF，ABFDAE.证明：延长AE交BF于点G.四边形ABCD是正方形，ABBC，BCFABE.又BECF，ABEBCF.CBFBAE.ABEEBGCBF90，ABEEBGBAE90.AGB90，即AEBF.(2)点E是OB的中点证明：延长AE交BF于H.四边形ABCD是正方形，ABBC，BCFABE.AEBF，AHB90.ABEEBHBAE

9、90.ABEEBHCBF90，CBFBAE.ABEBCF.BECF.BEOF，BEOC.OBOC，E是OB的中点between the requirements. And in the encounter the tedious work, complex things, sometimes take the method of escape, that bridge the boat to the bridge straight , not their own force please find countermeasures, but to wait for the self, the l

10、ack of a spirit of hard research.Two, the cause of the problemThrough the analysis, I feel that there are many reasons for these problems.One idea is obsolete. The failure to do system research on the essence of reading the original, palm and master the theory of quality read without thorough understanding, has not been effectively6