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1、作业练习学科初中数学年级九年级学期春季课题相似三角形中的分类讨论问题一温州卷第24题教科书书名:九年级数学教材出版社:浙江教育出版社达版社出版日期:2014年6月作业练习课程基本信息1 .如图1,以AA5C的边4?为直径作OO交3c于点。,连接4),点石为45上一点(不与 端点重合),连接CE,作。b_LC于点尸,延长上交AC于点交84的延长线于点G, ZBGD=ZACE.(1)求证:BG是。的切线;(2)求证:=; AB AE2(3)如图2,延长 CE 交 AB 于点 H,若 HE = 4, ZACH = ZBCH , sin ZBGD =-,求 6c 5的长.2 .已知:AABC内接于 O
2、 ,弦AD交3c于点石,OA为。的半径,ZOAC=ZBAD.(1)如图,求证:ADA.BC;(2)如图,弦族,AC于点H,交AD于点G,求证:GH = FH ;39(3)在(2)的条件下,NBM+4NC4D = 180。,连接 C5、EH ,如图,若 CF = 5, BC = ,求线段的长.图图参考答案1.【解答】解:(1) ZBGD = ZACE,且 NAMG = NCMF, /. 180O - Z8GD-ZAMG = 180O- ZACE-ZCMF,即 Z.GAM = ZCFM ,DFA.CE, .ZGAM = ZCFM=9Q , :.OAA.BG, .BG是 。的切线;(2) AC 为O
3、 直径,.ZAZ)C = ZAZ)3 = 90。, OALBG, ZB = 900 - ZACB = ZDAC,Ar1 An/. AADBCDA,一 = , .ABAD = ACBD, AB BDZB = ZDAC, ZBGD = ZACE, /. ABGDAACE,BG _BD:.ACBD = BGAE:,ABAD = BGAEBG AD(3)设CH交OO于N ,连接4V,如图:AC为 _O直径,.ZADC = 90 = ZANC,DFCE, /.ZFCD = 90-ZFDC = ZEDF ,AN!IDG, ZBGD = ZHAN, ZNAD = ZADG, ZACH = /BCH, ZBG
4、D= ZACE, /. ZACH = /BCH = ZEDF = Z.BGD = ZHAN = ZNAD ,在 AA/77V 和 AAE2V 中,ZHAN =4 NAD = 2,则 Na4T)= 180。4,ZCBF = 900-ZBCH = 900-ZACE = ZCAD,ZCBF = a,ABAC = ACAD + ABAD = a + (180-4a) = 180-36r, .ZBFC = ZBAC = lS00-3a, NBCF = 180。一 ZCBF - ZBFC = 180。一。(180。- 3。)= la ,FM = FC, /. ZCMF = ZBCF = 2a , ZBFM
5、 = ZCMF - ZCBF = a = ZCBF ,3914.BM = FM = FC = 5, :.CM = BC-BM =5 =一,55MN = CN = , :.FN = 4fM2 -MN2 =, BN = BM + MN = , 555BfZbM + FN? =8, /CNF = /CHF = 90。,.H、N、。、方四点共圆, :./BNH = /BFC ,而 ZHBN = NCBF , /. ABHNABCF ,32BH BN niI BH VBC BF 398y156 44FN 3GH = FH = BF BH = 8= tana =,BN 4AB = J AH? + BH2=卫, 3”四点共圆,./CHE = /CBA,25 255 GH 176FN 3/. AH =, sin a =,tana 75BF 5ZAHB = ZAEB = 9QP , .A、B、而 ZACB = ZECH ,. ACHEsACBA ,.叽工sin/,.EH = 4.AB AC BF 5
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