初三数学综合题

初三数学综合题 中国领先的高端教育连锁集团 初三数学补充综合题 1(已知:如图八，在,ABC中，AD,BC，D点为垂足，AC,BE，E点为垂足，M点位AB边的中点，联结ME、MD、ED( A ,MED,BMD (1)求证:与都是等腰三角形; )，EMD,2，DAC (2)求证:( E 图M 八 ) B C D 1 中国领先的高端教育连锁集团 3AC,BD,102( 如图十，C在射线BM上，在平行四边形ABCD中，，，tan，CAD,4对角线AC与BD相交于O点(在射线BM上截取一点E，使OC,CE，联结OE，与边CD相交于点F( (1)求CF的长; OC,CE(2)在没有“”的条件下，联结DE、AE，AE与对角线BD相交于P点，若,ADE为等腰三角形，请求出DP的长( A D ) 备 用 O 图 ) B C M 2 中国领先的高端教育连锁集团 3(已知?MON = 60?，射线OT 是?MON的平分线，点P是射线OT上的一个动点，射线PB交射线ON于点B( (1)如图十一，若射线PB绕点P顺时针旋转120?后与射线OM交于A，求证:PA = PB; 3(2)在(1)的条件下，若点C是AB与OP的交点，且满足PC =PB，求:?POB2与?PBC的面积之比; (3)当OB = 2时，射线PB绕点P顺时针旋转120?后与直线OM交于点A(点A不与点O重合)，直线PA交射线ON于点D，且满足(请求出OP的长( ，PBD,，ABO M M M ))) 备备T T T 图P 用用 十图图一 A 一二)C )) O N O O B N N 3 中国领先的高端教育连锁集团 4. 已知:梯形ABCD中，AD//BC，AB?BC，AD=3，AB=4，BC=6(如图)，点P为射线DC上的动点(不与D和C重合)，AP交BD于点E，连BP( (1)求tanC的值; (2)当点P在线段DC上时，如果?ADE与?BPC相似，求DP的长; S,PAD(3)设，试用的代数式示的值，并写出相应的的取值范围( DPx,xxS,PBC D A E P C B (图4) 4 中国领先的高端教育连锁集团 1(:(1)?M为AB边的中点，AD?BC， BE?AC， 11 ?，„„„„„„„„„„„„„„„„„„„„„(2分) MEAB,MDAB,22 ?ME=MD„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„(3分) ??MED为等腰三角形„„„„„„„„„„„„„„„„„„„„„„„„(5分) 1 (2)? MEABMA,,2 ??MAE=?MEA„„„„„„„„„„„„„„„„„„„„„„„„„„ (6分) ??BME=2?MAE„„„„„„„„„„„„„„„„„„„„„„„„„„(7分) 1同理可得: MDABMA,,2 ??MAD=?MDA„„„„„„„„„„„„„„„„„„„„„„„„„„ (8分) ??BMD=2?MAD„„„„„„„„„„„„„„„„„„„„„„„„„„(9分) ??EMD=?BME,?BMD =2?MAE,2?MAD=2?DAC„„„„„„„„„„„„„„„„„(10分) 2(解:(1)?ABCD为平行四边形且AC=BD ?ABCD为矩形„„„„„„„„„„„„„„„„„„„„„„„„„„„„(1分) ??ACD=90? CD3在RT?CAD中，tan?CAD= ,AD4 设CD=3k，AD=4k ?(3k)?+(4k)?=10? 解得k=2 ?CD=3k=6 „„„„„„„„„„„„„„„„„„„„„„„„„„„„„(2分) (?)当E点在BC的延长线上时， 过O作OG?BC于G„„„„„„„„„„„„„„„„„„„„„„„„„(3分) OGBO1,,? ?OG=3 CDBD2 BGBO1,,同理可得:，即BG=GC=4 GCOD1 1又? OC,CE,AC,52 CF5CFCE,?, ? OGEG35，4 5解得CF,„„„„„„„„„„„„„„„„„„„„„„„„„„„„„(4分) 3 (?)当E点在边BC上时，易证F在CD的延长线上，与题意不符，舍去„„(6分) (注:若有考生求出该情况下CF的长，但没有舍去此解，扣1分) ((( ,ADE(2)若为等腰三角形， AD,ED,8(?)(交于BC的延长线上) 2222CE,DE-DC,8-6,27由勾股定理可得:„„„„„„„„„(7分) 5 中国领先的高端教育连锁集团 ?AD?BE 82747BEBP，，令? ,,,,,,a84ADPD 4a，7a，4a?BP+PD=BD=10= 10(8,7)解得 a,57 40(8,7)320,407?„„„„„„„„„„„„„„„„(8分) PD,4a,,5757(?)AD,ED,8(交于边BC) 82747BPBE,,令同理可得: ,,,,,,a84PDAD BP，PD,BD,10,4a,7a，4a? 10(8，7)解得a, 57 40(8，7)320，407PD,4a,,?„„„„„„„„„„„„„„„„(9分) 5757 AE,ED(?) ,AEB,,DEC易证: 1? BE,EC,BC,42 BP1BP1?同理可得:,，则 ,BD3103 1020?，PD=„„„„„„„„„„„„„„„„„„„„„„„„(10分)BP,33 AE,AD,8(?) 22BE,8,6,27? 7BEBP令?同理可得: ,,,,,a4ADPD4a，7a,10 10(4,7)a,9 6 中国领先的高端教育连锁集团 160,407?„„„„„„„„„„„„„„„„„„„„„„(11分) PD,4a,9 320,407320，40720,ADE?综上所述，若为等腰三角形，或PD,或或57573160,407„„„„„„„„„„„„„„„„„„„„„„„„„„„„(12分) 9 (注:若考生只详细写出一种情况，其余几种均用了同理，只要正确，也给满分) (((( 3(解:(1)证明:作PF?OM于F，作PG?ON于G„„„„„„„„„„„„(1分) ?OP平分?MON ?PF =PG„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„(2分) ??MON = 60? ??FPG = 360?– 60?– 90?– 90?= 120?„„„„„„„„„„„„„„„„„„(3分) 又??APB =120? ??APF = ?BPG ??PAF??PBG„„„„„„„„„„„„„„„„„„„„„„„„„„„(4分) ?PA = PB„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„(5分) (2)由(1)得:PA = PB，?APB =120? ??PAB = ?PBA = 30?„„„„„„„„„„„„„„„„„„„„„„„„(6分) ??MON = 60?，OP平分?MON ??TON = 30?„„„„„„„„„„„„„„„„„„„„„„„„„„„„(7分) ??POB = ?PBC„„„„„„„„„„„„„„„„„„„„„„„„„„„(8分) 又?BPO = ?OPB ??POB??PBC„„„„„„„„„„„„„„„„„„„„„„„„„„„(9分) SPBPB4,22POB,(),(),? SPC33,PBCPB2 ??POB与?PBC的面积之比为4?3„„„„„„„„„„„„„„„„„„(10分) 7 中国领先的高端教育连锁集团 (3)? 当点A在射线OM上时(如图乙1)，易求得:?BPD = ?BOA = 60? ?，PBD,，ABO，而?PBA = 30?，??OBA = ?PBD = 75? 作BE?OT于E ??NOT = 30?，OB = 2 3?BE =1，OE = ，?OBE = 60? ??EBP = ?EPB = 45? ?PE = BE =1 3?OP = OE + PE =+ 1„„„„„„„„„„„„„„„„„„„„„„„(12分) ? 当点A在射线OM的反向延长线上时(如图乙2) 此时?AOB = ?DPB = 120? ，PBD,，ABO?，而?PBA = 30?，??OBA = ?PBD = 15? 作BE?OT于E 3??NOT = 30?，OB = 2，?BE =1，OE = ，?OBE = 60? ??EBP = ?EPB = 45? ?PE = BE =1 3?OP =,1„„„„„„„„„„„„„„„„„„„„„„„„„„„„(14分) OP,3，1或3,1OB,2?综上所述，当时， (注:若考生直接写出结果，只给一半的分数) ((((((((((((( M M A T T E P P E O D N B D O N B A 图乙2 图乙1 8 中国领先的高端教育连锁集团 已知:梯形ABCD中，AD//BC，AB?BC，AD=3，AB=4，BC=6(如图)，点P为射线DC上的动点(不与D和C重合)，AP交BD于点E，连BP( (1)求tanC的值; (2)当点P在线段DC上时，如果?ADE与?BPC相似，求DP的长; S,PAD(3)设，试用的代数式表示的值，并写出相应的的取值范围( DPx,xxS,PBC D A 解:(1)过D作DH?BC，---------------------------------------1分 E AD//BC,AB?BC,?AB?AD, ?ABHD为矩形，?DH=AB=4,BH=AD=3, ? ?CH=3-----1分 P DH4?DH?BC，?----------------------------------1分，1分 C tanC,,B HC3(图14) (2)?AD=3,AB=4, AD//BC,AB?BC,?BD=5,而DH=4,HC=3, DH?BC, ?DC=5 , ?DC=BD,??DBC=?C ?AD//BC, ??ADB=?DBC, ??ADB=?C-------------------------------1分 方法1:??ADE与?BPC相似，又?ADB=?C ??DAP=?PBC或?DAP=?BPC----------------------------------------2分 当?DAP=?PBC时， ?AD//BC, ,AB?BC,??PAB=?PBA，?PA=PB， DPAN取AB中点N，连PN，?PN?AB, ?PN//AD//BC, ?, ,PCNB 5?DP=PC,即DP=---------------------------------------------------1分 2 当?DAP=?BPC时，在BP的延长线上去点Q，则?BPC=?DPQ， ??DPQ=?DAP+?ADP,?DAP，??DAP=?BPC不可能-------------------1分 5?DP=(注:只要说明?DAP=?BPC时不可能，就得1分) 2 方法2:??ADE与?BPC相似，又?ADB=?C ADCPADBC?或-------------------------------------------2分 ,,DEBCDECP 延长AP交BC的延长线于点M,设DP=x，则PC=5-x ADDP3x153,x?AD//BC, ?,即，? ,,CM,CMPCCMx5,x 9 中国领先的高端教育连锁集团 3DEADDE5x,即,? 又?AD//BC, ?,,DE,153,x5,DEBMBE25x，6，x 35,x3622511900xx，，,25250xx，,,?或，即或 ,,5x5x65,x 25x，25x， 2511900xx，，,由得无解;------------------------------------------1分 55225250xx，,,由得(舍)即得DP=------------------1分 xx,,,,51222 (3)当点P在线段CD上时，过,做,,?,,分别交直线,,和直线,,于,和,， SADPNDPxx,31PAD?()------------1分 05,,x,,,,,,SBCPMPCxx,,,625102PBC 当点P在DC的延长线上时，过,做,,?,,分别交直线,,和直线,,于,和,， SADPNDPxx,31PADx,5()------------------,分 ,,,,,,SBCPMPCxx,,,625210PBC , A A D D , E E P C M B C B , P 10