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(完整word版)初中数学圆的辅助线八种作法.pdf

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1、.中考数学圆的辅助线在平面几何中，与圆有关的许多题目需要添加辅助线来解决。百思不得其解的题目，添上合适的辅助线，问题就会迎刃而解，思路畅通，从而有效地培养学生的创造性思维。添加辅助线的方法有很多，本文只通过分析探索归纳几种圆中常见的辅助线的作法。下面以几道题目为例加以说明。1.有弦，可作弦心距在解决与弦、弧有关的问题时，常常需要作出弦心距、半径等辅助线，以便应用于垂径定理和勾股定理解决问题。例1如图 1,O 的弦 AB、CD 相交于点 P，且 AC=BD。求证：PO 平分 APD。分析 1：由等弦 AC=BD 可得出等弧=进一步得出=，从而可证等弦AB=CD，由同圆中等弦上的弦

2、心距相等且分别垂直于它们所对应的弦，因此可作辅助线OEAB，OFCD，易证 OPE OPF，得出 PO 平分 APD。证法 1：作 OEAB 于 E，OFCD 于 F AC=BD=AB=CD=OE=OF OEP=OFP=90=OPE OPF 0OP=OP=OPE=OPF=PO 平分 APD 分析 2：如图 1-1，欲证 PO 平分 APD，即证AB(BD，(CD(D C B P OA E F P B 图 1 AC(AC(BD(AB(CD(.OPA=OPD，可把 OPA 与 OPD 构造在两个三角形中，证三角形全等，于是不妨作辅助线即半径 OA，OD，因此易证 ACP DBP，得 AP=DP，从

3、而易证 OPA OPD。证法 2：连结 OA，OD。CAP=BDP APC=DPB=ACP DBP AC=BD=AP=DP OA=OD=OPA OPD=OPA=OPD=PO 平分 APD OP=OP 2.有直径，可作直径上的圆周角对于关系到直径的有关问题时，可作直径上的圆周角，以便利用直径所对的圆周角是直角这个性质。例 2 如图 2，在 ABC 中，AB=AC，以 AB 为直径作 O 交 BC 于点 D，过 D 作 O 的切线 DM 交 AC 于 M。求证DMAC。分析：由 AB 是直径，很自然想到其所B D C M A O.A 2 1 图 2 D C B P OA P B 图 1-1 文档编

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11、2+ADM=AMD=ADB=Rt =DM AC 说明，由直径及等腰三角形想到作直径上的圆周角。3.当圆中有切线常连结过切点的半径或过切点的弦例 3 如图 3,AB 是 O 的直径，点 D 在 AB 的延长线上，BD=OB，DC 切 O 于C 点。求 A 的度数。分析：由过切点的半径垂直于切线，于是可作辅助线即半径OC，得 Rt，再由解直角三角形可得COB 的度数，从而可求 A 的度数。解：连结 OC。DC 切 O 于 C=OCD=90 OC=OB=BD=A=1/2 COB=30 说明，由过切点的半径垂直于切线想到连结半径。例 4 如图 4，已知 ABC 中，1=2，=1=2=COS COD=O

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19、个角分别等于1 和 2，故易证 EF/BC。证明连结 DE。BC切 O 于 D=BDE=1 2=DEF=BDE=DEF=EF/BC 1=2 说明，由有切线且在同圆中等弧所对的圆周角相等想到连结弦。4.当两圆相切，可作公切线或连心线例 5 已知：如图5，O1与 O2外切于点 P，过 P 点作两条直线分别交O1与O2于点 A、B、C、D。求证PB?PC=PA?PD。分析：欲证PB?PC=PA?PD，即证 PAPB=PCPD，由此可作辅助线AC、BD，并证 AC/DB，要证平行，需证一对内错角相等，如C=D，然后考虑到这两个角分别与弦切角有关，进而再作辅助线即两圆公切线MN，从而问题迎刃而解。A C

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25、Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F

26、9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9.证明连结 AC、BD，过 P 点作两圆的内公切线MN=APM=C，BPN=D APM=BPN=AC/DB=PAPB=PCPD=PB?PC=PA?PD 说明，由需证弦平行且弦切角等于其所夹

27、弧对的圆周角想到作公切线和作弦。例6已知：如图6，O1与 O2内切于点 T，经过切点 T 的直线与 O1与 O2分别相交于点A 和 B。求证TATB=O1AO2B。分析：欲证TATB=O1AO2B，可考虑证这四条线段所在的三角形相似，即证 TO1A TO2B，于是只需连结O2O1，并延长，必过切点，则产生 TO1A 和 TO2B，由 1=2=T，则 O1A/O2B，易证线段比相等。证明连结并延长O2O1O1 和 O2内切于点 T O1A=O1T=1=T O2T=O2B=2=T=TO1A TO2B=TA TB=O1AO2B 说明，由连心线必过切点可构造三角形证全等想到作连心线。T B A O1

28、O21 2 图 6=C=D=O2O1必过切点 T=1=2=O1A/O2B文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10

29、HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 Z

30、B5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编

31、码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I1

32、0 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3

33、 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文

34、档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9.5当两圆相交，可作公共弦或连心线。例 7 如图 7，O1与 O2相交于 A、B 两点，过 A 点作 O2的切线交 O1于点 C，直线 CB交 O2于点 D，DA 延长线交 O1 于点 E，连

35、结 CE。求证CA=CE。分析：欲证CA=CE，考虑在三角形中证它们所对的角相等，即E=CAE，又由 DAF=CAE，想到弦切角 DAF 与所夹弧对的圆周角相等，故需作辅助线：公共弦 AB，得 E=DBA，易证 CA=CE。证明连结 AB。CA 切 O2于 A=DAF=DBA 四边形 ABCE内接于 O1=E=DBA DAF=CAE=E=CAE=CA=CE 说明，由两圆相交及用到弦切角和圆内接四边形想到作公共弦。C D E M NG A B O2O1F 图 8 F E B C A O1O2.图 7 D 文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:

36、CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10

37、HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 Z

38、B5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编

39、码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I1

40、0 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3

41、 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文

42、档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9.例8如图 8，在梯形 ABCD 中，以两腰AD、BC 分别为直径的两个圆相交于M、N 两点，过 M、N 的直线与梯形上、下底交于E、F。求证：MNAB。分析：因为MN 是公共弦，若作辅助线O1O2，必有 MN O1O2，再由 O1O2是梯形的中位线，得O1O2/AB，从而易证MN AB。证明连结 O1O2交 EF于 G=MN O1O2。DO1=O1A，

43、CO2=O2B=O1O2是梯形 ABCD 的中位线=O1O2/AB=EFA=EGO1=Rt=MN AB 说明，由两圆相交连心线垂直于公共弦想到作连心线。6有半圆，可作整圆例 9 如图 9，BC为 O 的直径，ADBC于 D，=,AD 交 BF于 E。求证AE=BE 分析：欲证AE=BE，可考虑在三角形中证这两边所对角相等。即ABF=BAE，再考虑证这两个圆周角所对的弧相等，故需补全O，可证=，故有=易证 AE=BE.证明补全 O，延长 AD 交 O 于 H，直径 BCAD=ABF=BAH=AE=BE 说明，由平分弦的直径必平分弦所对的弧想到补全圆。7相交两圆中至少有一个圆经过另一个圆的圆心，遇

44、到这类问题，常用的辅助线是连结过交点的半径例 10 如图 10，O1与 O2相交于A、B 两点，且O2在 O1上，点 P 在 O1上，点 Q 在 O2上，若 APB=40，求 AQB 的度数。BA(AF (BA(BH(AF(BH(F A B D O.H E C 图 9 BA(BH(AF，(BH(BA=AF，(文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H

45、9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4

46、K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ

47、4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX

48、8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5

49、K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:

50、CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10 HX8H9O5P4Y3 ZB5K4K8M9F9文档编码:CJ4T1Z7A9I10

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本文标题：(完整word版)初中数学圆的辅助线八种作法.pdf
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