(完整word版)初中数学圆的辅助线八种作法.pdf

.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: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 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 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: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 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.对的圆周角是直角。于是可连结AD，得 ADB=Rt,又由等腰三角形性质可得1=2，再由弦切角的性质可得ADM=B，故易证 AMD=ADB=90 ，从而 DMAC。证明连结 AD。AB 为 O 的直径=ADB=Rt AB=AC DM 切 O 于 D=ADM=B=1+B=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=OC/OD=1/2=COB=60 D A O B C.图 3 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: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 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 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: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 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.圆 O 过 A、D 两点，且与BC切于 D 点。求证EF/BC。分析：欲证EF/BC，可找同位角或内错角是否相等，显然同位角相等不易证，于是可连结DE，得一对内错角BDE 与 DEF，由圆的性质可知这两个角分别等于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 N B D M P O1O2.E D C F O 1 2 A B 图 4 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: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 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 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: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 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 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 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 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: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 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 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 ZB5K4K8M9F9.5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: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 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 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: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 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 ZB5K4K8M9F9文档编码: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，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 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 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: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 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 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文档编码:CJ4T1Z7A9I10