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-1-1.2.2圆周角定理ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航1.掌握圆周角的概念.2.理解圆周角定理及其三个推论,并能解决有关问题.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航1.圆周角的概念从☉O上任一点P引两条分别与该圆相交于点A和点B的射线PA,PB,叫做∠APB所对的弧,∠APB叫做所对的圆周角.𝐴𝐵𝐴𝐵ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航2.圆周角定理文字语言圆周角的度数等于它所对弧的度数的一半图形语言符号语言在☉O中,BC所对的圆周角是∠BAC,则有∠BAC=12BC的度数作用确定同一个圆中弧长与角的大小关系ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航【做一做1】如图,在☉O中,∠BAC=25°,则的度数等于()A.25°B.50°C.30°D.12.5°答案:B解析:根据圆周角定理,得𝐵𝐶的度数=2∠BAC=50°.𝐵𝐶ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航3.圆周角定理的推论(1)推论1:直径(或半圆)所对的圆周角都是直角.(2)推论2:同弧或等弧所对的圆周角相等.(3)推论3:等于直角的圆周角所对的弦是圆的直径.名师点拨1.“相等的圆周角所对的弧也相等”的前提条件是“在同圆或等圆中”.2.由弦相等推出弧相等时,这里的弧要求同是优弧或同是劣弧,一般选劣弧.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航【做一做2-1】如图,在☉O中,∠BAC=60°,则∠BDC等于()A.30°B.45°C.60°D.75°解析:∠BDC=∠BAC=60°.答案:CZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航【做一做2-2】如图,AB是☉O的直径,C是𝐴𝐵上的一点,且AC=4,BC=3,则☉O的半径r等于()A.52B.5C.10D.不确定解析:∵AB是☉O的直径,∴∠ACB=90°.∴AB=𝐴𝐶2+𝐵𝐶2=42+32=5.∴2r=AB=5.∴r=52.答案:AZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航相等的圆周角所对的弧不一定相等剖析“相等的圆周角所对的弧相等”是在“同圆或等圆中”这一大前提下成立的.如图,若AB∥DG,则∠BAC=∠EDF,但𝐵𝐶≠𝐸𝐹.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型一求线段的长【例1】如图,△ABC的三个顶点都在☉O上,∠BAC的平分线与BC边和☉O分别交于点D,E.(1)指出图中相似的三角形,并说明理由;(2)若EC=4,DE=2,求AD的长.分析(1)本题要用三角形相似的判定定理,而其中角的条件由同弧所对的圆周角相等得出;(2)要求线段长度,先由三角形相似得线段成比例,再求其长度.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三解:(1)∵AE平分∠BAC,∴∠BAD=∠EAC.又∠B=∠E,∴△ABD∽△AEC.∵∠B=∠E,∠BAE=∠BCE,∴△ABD∽△CED,△AEC∽△CED.∴CE2=ED·AE.∴16=2×AE.∴AE=8.∴AD=AE-DE=6.反思求圆中线段的长时,常先利用圆周角定理及其推论得到相似三角形,从而得到成比例线段,再列方程求得线段长.(2)∵△CED∽△AEC,∴𝐶𝐸𝐴𝐸=𝐸𝐷𝐸𝐶.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型二证明线段相等【例2】如图,BC为圆O的直径,AD⊥BC,,BF和AD相交于点E.求证:AE=BE.分析要证明AE=BE,只需在△ABE中证明∠ABE=∠EAB,而要证明这两个角相等,只需借助∠ACB即可.证明∵BC是☉O的直径,∴∠BAC为直角.又AD⊥BC,∴Rt△BDA∽Rt△BAC.∴∠BAD=∠ACB.∵,∴∠FBA=∠ACB.∴∠BAD=∠FBA.∴△ABE为等腰三角形.∴AE=BE.𝐴𝐹=𝐴𝐵∵𝐴𝐵=𝐴𝐹ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三反思1.有关圆的题目中,圆周角与它所对的弧经常相互转化,即欲证圆周角相等,可转化为证明它们所对的弧相等;要证线段相等可以转化为证明它们所对的弧相等,这是证明圆中线段相等的常见策略.2.若已知条件中出现直径,则常用“直径所对的圆周角为直角”这一性质解决问题.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三题型三易错辨析易错点:误认为同弦或等弦所对的圆周角相等【例3】如图,∠BAD=75°,求∠BCD.错解:∵∠BAD和∠BCD所对的弦都是BD,∴∠BAD=∠BCD.∴∠BCD=75°.错因分析错解中,没有注意到圆周角∠BAD和∠BCD所对的弧不相等,导致得到错误的结论∠BAD=∠BCD.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航题型一题型二题型三正解:∠BAD是𝐵𝐶𝐷所对的圆周角,∠BCD是𝐵𝐴𝐷所对的圆周角,则𝐵𝐶𝐷所对的圆心角为2×75°=150°.∵𝐵𝐶𝐷和𝐵𝐴𝐷所对的圆心角的和是周角360°,∴𝐵𝐴𝐷所对的圆心角是360°-150°=210°,∴𝐵𝐴𝐷所对的圆周角∠BCD=12×210°=105°.反思在同圆中,同弦或等弦所对的圆周角不一定相等.当弦是直径时,同弦或等弦所对的圆周角相等,都等于90°;当弦不是直径时,该弦将圆周分成两条弧:优弧和劣弧,若圆周角的顶点同在优弧上或同在劣弧上,同弦或等弦所对的圆周角相等;若一个圆周角的顶点在优弧上,另一个圆周角的顶点在劣弧上,则同弦或等弦所对的圆周角不相等,它们互补.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123451.如图,弦AC与BD相交于圆内一点P,且AB=10,CD=5,BP=8,则PC=.解析:∵∠A=∠D,∠C=∠B,∴△ABP∽△DCP.答案:4∴𝐴𝐵𝐶𝐷=𝐵𝑃𝑃𝐶.∴105=8𝑃𝐶,解得PC=4.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123452.如图,AC是☉O的直径,B是圆上一点,∠ABC的平分线与☉O相交于点D,已知BC=1,AB=,则AD=__________.3ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航12345解析:如图,连接OD,由于AC是☉O的直径,则∠ABC=90°.又BC=1,AB=3,则AC=𝐴𝐵2+𝐵𝐶2=12+(3)2=2,所以OA=OD=12AC=1.又∠AOD=2∠ABD=∠ABC=90°,故△AOD是等腰直角三角形,则AD=2OA=2×1=2.答案:2ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123453.如图,AB是☉O的直径,BD是☉O的弦,延长BD到点C,使AC=AB.求证:BD=DC.证明如图,连接AD.∵AB是☉O的直径,∴∠ADB=90°,即AD⊥BC.又AC=AB,∴BD=CD.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123454.如图,已知AD是△ABC的高,AE是△ABC的外接圆的直径,AD的延长线交外接圆于点F.分析转化为证明∠BAE=∠FAC,再转化为证明△ABE∽△ADC.证明∵AE是直径,∴∠ABE=90°.又∠ADC=90°,∴∠ADC=∠ABE.又∠AEB=∠DCA,∴△ABE∽△ADC.∴∠BAE=∠FAC.求证:𝐵𝐸=𝐹𝐶.∴𝐵𝐸=𝐹𝐶.ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航123455.如图,△ABC的角平分线AD的延长线交它的外接圆于点E.(1)证明:△ABE∽△ADC;(2)若△ABC的面积S=AD·AE,求∠BAC的大小.分析(1)证明这两个三角形的两个角对应相等;(2)利用(1)的结论和三角形面积公式的正弦形式,转化为求sin∠BAC.12ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航12345(1)证明∵AD平分∠BAC,∴∠BAE=∠CAD.又∠AEB与∠ACD是同弧所对的圆周角,∴∠AEB=∠ACD.∴△ABE∽△ADC.∴AB·ACsin∠BAC=AD·AE.∴sin∠BAC=1.又∠BAC为三角形的内角,∴∠BAC=90°.(2)解:∵△ABE∽△ADC,∴𝐴𝐵𝐴𝐸=𝐴𝐷𝐴𝐶,即AB·AC=AD·AE.又S=12AB·ACsin∠BAC,且S=12AD·AE,ZHISHISHULI知识梳理ZHONGNANJVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLITOUXI典例透析MUBIAODAOHANG目标导航