题目 #21051

【讲解】 2026-06-22 (1)EF与 O相切；理由如下： (1)\(\frac{四边形}{AED}\frac{AEGF}{DFG}\)是 O的内接四边形，  =  = 60， 如图，连接OB，  CB = CD,CDB = CBD, 故答案为：；60   = 90,\(\frac{AOC}{ADO}\)(2)连接EF，如图①所示：  +  = 90, 四边形FGEA是 O的内接四边形，\(\frac{OAD}{CDB}\) = ， 又 ADO = ，\(\frac{DGF}{GDF} \frac{DAE}{ADE}\)又 = ，DFG  DEA,  =  = CBD,\(\frac{ADO}{CBD} \frac{CDB}{OAD}\)当DEA为等腰三角形时，  +  = 90,  OA = OB, = , DFG为等腰三角形，\(\frac{OAD}{OBD} \frac{OBD}{CBO}\) 四边形ABCD是矩形，AB = 4, BC = 6，  +  =  = 90,\(\frac{CBD}{OB}\)CD = AB = 4, AD = = 6, 即 ⊥ BC,\(\frac{BC}{BCD}\) = ABC =  =  = 90,\(\frac{BAD}{O} \frac{ADC}{ABE} \frac{OB}{EF}\) 为半径，  是 的外接圆， = 90，  与 O相切；\(\frac{ABE}{O}\) AE是  的直径, (2)如(1)图， = 90，\(\frac{CBO}{3}\)AFE = 90,\(\frac{O}{OA}\) 的半径为，  = OB = 3,  = 180− = 180 − 90 = 90，\(\frac{DFE}{CDF} \frac{AFE}{DFE}\) AOC = 90,tan OAD =\(\frac{1}{3}\) = DCE =  = 90, 四边形 是矩形，\(\frac{DCEF}{CEEF}\)  =\(\frac{DO}{AO} = \frac{1}{3},OD = 1\)DF =, = CD = 4,\(\frac{tan}{BC} \frac{OAD}{CD} \frac{AED}{AE}\)若 为等腰三角形，分三种情况： 设 = = x, = DE时，  = 90,\(\frac{AFE}{DF}\)①当 CO = + DO = x +1，\(\frac{CD}{RtBOC}\)⊥  AF = =\(\frac{1}{2}\)AD = 3,  在  中，\(\frac{EF}{CE} \frac{AD}{DF}CO^{2}\)= +\(OB^{2}\)，\(CB^{2}_{2}\) = = 3; （x +1） =\(3^{2}\)+\(x^{2}\)， = AD = 6时，\(\frac{AE}{RtABE}\)②当 解得：x = 4, 在  中，由勾股定理得： BC = 4. = =\(\sqrt{6^{2}−4^{2}} = 2 \sqrt{5}\)，\(\frac{BE}{CE} \frac{\sqrt{AE^{2}−AB^{2}}}{BC}\) = − = 6 −\(2 \sqrt{5}\)；\(\frac{CE}{DA}\)③当DE = = 6时，在RtDCE中， 由勾股定理得： CE =\(\sqrt{DE^{2}−CD^{2}} = \sqrt{6^{2}−4^{2}} = 2 \sqrt{5}\)； 综上所述，当BE的长为或36 −\(2 \sqrt{5}\)或\(2 \sqrt{5}\)时， DFG为等腰三角形， 故答案为：或36 −\(2 \sqrt{5}\)或2 5; 2026-06-22 (3)过作OOH ⊥ CD于点， H 如图②所示： 则OH∥AD∥CE，\(\frac{四边形}{ABC}\frac{ABCD}{90}\) 是矩形，  = ，  为 \(\frac{AE}{OA}\)O的直径，  =\(\frac{OE}{是梯形}\)\(\frac{OH}{OH}\)ADCE的中位线，  = 1 (+)\(2\frac{AD}{AD} \frac{CE}{CE}\)2 = +,\(\frac{OH}{O}\)相切， 为切点,\(\frac{CD}{OA}\)H  与  =\(\frac{OH}{AE}\),  = 2OH = AD + CE = 6 + CE, 在Rt ABE中由勾股定理得, : + =,\(\frac{AB^{2}}{4^{2}} \frac{BE^{2}}{6}AE^{2}_{2}\)即 +（− CE）（=6 + CE）2, 解得 : CE=\(\frac{2}{3}\). 2026-06-22