1、 , 4 I.ft I T * - eI/ P Jz I I r Q :L与3尸的关系,并证明你的结论;求HE:MB = MDD12.如图,点P是/!BC内一点,EF分别是边Ae,BC上的两点,连接PE, PF,且 PE = PF,点 D 为 AC延长线上一点,连接PD,且DE=BF,ZAEP+ ZBFP = 180。.(1)求证:ZXDfP 三 ABFP;(2)已知 AB = AE+BF,若 ZACB = SOo9 求 ZAP3 的度数3.填空题13如图将厶ABC绕点C按顺时针方向旋转至AfB,C .使点A落在BC的延长线上已知ZA = 27。,ZB = 40,则ZAeBz= 度.14.如图
2、,AC = BC.请你添加一对边或一对角相等的条件,使AD = 3你所添加的条件是 15.如图,O是AABC内一点,且O到三边AB. BC, CA的距离OF = OD = OE ,若ZBAC = 70,16已知:如图在长方形ABCD中,AB = AD = 6.延长BC到点E使CE = 2,连接DE,动点P从点B出发,以每秒2个单位的速度沿BCCDD4向终点A运动,设点P的运动时间为/ 秒,当F的值为 秒时,MBP和ADCE全等.17如图所示,已知EA丄AB. BCll EAyEA = AB = IBC, D为AB的中点,那么下列结论:DE = AC ,DE丄4C,ZEAF = ZADF ,(4
3、) ZC = ZADF中正确的有_(填序号)参考答案1.答案:C解析:根据题意,将周长为16Cm的AABC沿BC方向平移2cm得到ADEF,. AD = CF = 2cm,BF = BC+ CF = BC+ 2, DF = AC 又48 + 80 + 4( = 1601,二四边形 ABFD 的周长为 AD +AB +BF+ DF = 2 +AB + BC + 2 +AC = 20(cm).2.答案:当厶 ABC 0PA 时,AP = BC = 6cm:当厶 ABC PQA 时,AP = AC = I2cm.故 选C.3.答案:B.8CB三MCB, .ZACB = ZA. ZACB - ZA =
4、 ZACBn - ZACB ,即 ZACA = ZBCg. ZBCB, = 320, .-.ZACAf 的度数为32。4.答案:T 点 O 为 AB CZ)的中点,.MO = BO, CO = DO.又. ZAoC = ADOB, .AOC三MOD(SAS),.AC = BD,二容器内径为15cm.故选D5.答案:解析:.BE丄CE, AD丄CE, .ZE = ZADC = 90。. ZEBC+ZBCE = 90。.ZBCEtZACD = QO-EBC = ZDCA.ZE = ZADC在 ZkCEB 和 AADC 中,ZEBC = ZDCA ,BC = AC.CEB 三ZADC( A4S),.
5、BE = DC = , CE = AD = 3.DE = EC-CD = 3- = 2.故选 B6.答案:如图,延长AD到点E,使DE = Ar,连接EC.AD是BC边上的中线,.BD = CD.BD = CD在 AABD 和 ZXECD 中, ZADB = AEDC , AD = DE. ABD =ECD(SAS), :. CE=AB.AB = 3, AC = 5, .5-3 A5+3 , 即22ADS, :AD4,故选 BE7.答案:A从题图上可以看出点M在ZAoB的平分线上,其他三点不在ZAOB的平分线上,所以点M到ZAoB两边的距离相等.故选A.8.答案:如图,连接AD,. BE丄AC
6、于点E,CF丄A3于点F,. ZAEB = ZAFC = 90o, ZDEC = ZDFB =90,在 MBE 和 AACF 中,ZBAE = ZFDB AAEB = ZAFC, .ABE ACF( AAS),故正确:. .AE = AFAB = AC. AB = AC,/. EC = BF ,住厶DEC 和厶DFB 中,ZEDC = ZFDBZDEC = ZDFB, . ADEC 三ADFB(AAS),故正确;. DE = DFEC = FB . DE丄AC, DF丄AB . D4平分ZCAB 故正确;9.答案:. ZA = 90, ZB = 25o, . ZC = 90- 25 = 65.
7、ABC与厶DEF全等,.ZE与ZB是对应角时,ZE = 25;ZE与Ze是对应角时,ZE = 65. ZE的度数是25或65.故选D.10.答案:AM = BKy在 ZXAMK 和 ABKN 中 J Z = ZB.AK = BN、. .AMK 三厶BKNZAMK = ZBKN.乙MKB = ZMKN + 乙NKB = ZA + ZAMK,. ZA = AMKN = 44. ZP = 180o-ZA-ZB = 9211.答案:解-.(I)DE = BF,且DE/BF.证明如下:. DE 丄 AC, BF 丄 AC,. ZDEC = ZBFA = 90o, . DEUBF. AE = CF,/.
8、ZAE+ EF = CF + EF、AF = CE.AB = CD,在RtAF和RtACDE中AF = CE,. RtAfiF RtCDE(HL), . DE = BF.ZDEM=ZBFM、(2)证明:在 DEM 和 ZBFM 中 J ZDME = ZBMF、DE = BF.,DEM 三ABFM (AAS), . MB = MD.12.答案:解:证明:. ZAEP + ZBFP = 180o, ZAEP + ZDEP = 180. ZDEP = ZBFP.又. DE = BF, PE = FP,.DEP 三 AEFP(SAS).込DEP 三厶BFP,:. PB = PDy ZD = ZFBP.
9、 AB = AE +BF = AE +DE = AD, AP = AP,. .APB 三AAPD(SSS),. ZD = ZABP = AFBP. APAD = ZPAB. ZACB = 80, /. ZCAB + ZCBA = 100. ZPAB + ZPBA = 50,/. ZAPB = 13013.答案:46 ABC绕点C按顺时针方向旋转至 ABfC,.ZABC AB,C, .ZA = ZAz, ZB = ZB. ZA = 27o, ZB = 40o, AZAZ = 27o, ZB= 40。,.ZAG4z = ZAZB = 27o+40o = 67o, ZBCB= ZA+ ZB= 27。
10、+40。= 67。. ZACBf = 180o -ZACA,- ZBCB = 180o-67 - 67 = 4614.答案:CZ) = CE (答案不唯一)因为 AC = BC, ZC = ZC,当 CD = CE时,AADC三厶BEe(SAS 所以 AD = BE.15.答案:125解IJf: .OF = OD = OE,:. OByOC 分别平分ZABC和 ZACB. / ZBAC = 70 ,. ZABC ZACB = 180 - 70 = 110. ZOBC + ZOCB =丄(ZABC + ZACB)=丄 Xlloo = 552 2. ABOC = 180。-(ZoBC + ZOCB
11、) = 180-55 = 12516.答案:1或7因为AB = CD、沁ABP三厶DCE时,BP = CE = 2.由题意得BP = 2 = 2,解得r = l.当厶ABPDCE时,AP = CE = 2,由题意得AP = I62/ = 2,解得7 = 7.所以当t的值为1或7时,AABP和(?全等.17.答案:.E4 丄 AB.:. ZEAD = 90又. BC/EA,:. ZABC = 90o, . ZEAD = ZABC = 90. D 为 43 的中点,EA = B = 2BC,. AD = BC, .,E4D ABC(SAS). DE = ACy ZC = ZADF ,故正确. BC/EA, :. ZC = ZEAF,:.ZEAF = ZADFy 故正确. ZEAD = 90o, . ZEAF + ZFAD = 90/. ZADF+ AFAD = ,即 ZAFD = 90.DE丄AC,故正确.综上可知,正确的结论有
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