1、第28讲 三角形的内切圆题一:在RtABC中,C = 90,AC = 12,BC = 5,则ABC的内切圆的半径是 .题二:RtABC中,C = 90,它的内切圆O分别与AB、BC、CA相切于D、E、F,且AD = 6,BD = 4,则O的半径是 .题三:如图,点E是ABC的内心,AE的延长线和ABC的外接圆相交于点D,连接BD、BE、CE,若CBD = 32,则BEC的度数为 122题四:已知ABC中A的平分线和外接圆O相交于点D,BE是O的切线,DFBE,DGBC,垂足分别为F、G求证:DF = DG第28讲 三角形的内切圆题一:2详解:如图,在RtABC,C = 90,AC = 12,B
2、C = 5,根据勾股定理得AB = 13,四边形OECF中,OE = OF,OEC = OFC = C = 90,四边形OECF是正方形,由切线长定理,得BD = BE,AD = AF,CE = CF,CE = CF =(AC+BCAB),即r =(12+513) = 2,故答案为2题二:2.详解:如图,O是RtABC的内切圆,四边形CEOF是正方形,AF = AD = 6,BE = BD = 4,设O的半径为r,则CE = CF = r,(4+r)2+(6+r)2 = (4+6)2,r = 2或r =12(舍掉),内切圆的半径是2题三:122详解:在O中,CBD = 32,CAD = 32,点E是ABC的内心,BAC = 64,EBC+ECB = (18064)2 = 58,BEC = 18058 = 122故答案为122题四:见详解详解:如图所示,连接BD,BE是圆O的切线,FBD = BAD,DBC = DAC,BAD = DAC,FBD = DBG,又DFBE,DGBC,DF = DG