题目 #21120

2.(1)见解析; (2)3√2. (1)连接OC, ∵AB是⊙O的直径, ∴∠ACB = 90∘, ∴∠A + ∠ABC = 90∘,∵OB = OC, ∴∠ABC = ∠OCB, ∵ ∠BCD = ∠A, ∴∠BCD + ∠OCB = 90∘,即∠OCD = 90∘, ∴OC ⊥CD. ∵OC为⊙O的半径, ∴CD是⊙O的切线. (2)∵ 点 B 是 AD 的中点, ∴BD = AB = 2OC, ∵OB = OC, ∴OD = OB +BD= 3OC, ∴ OC\(1 _{OD}\)= _{3}, ∵BE ⊥AD, ∴ ∠DBE = 90∘, 又∵∠OCD = 90∘,∴sinD = BE OC\(1 _{DE}\)= _{OD}= _{3}, ∴DE = 3BE = 9, 在Rt △DBE中,BD=\(\sqrt{DE^{2}}\)−\(BE^{2}\)=\(\sqrt{9^{2}}\)−\(3^{2}\)= 6√2. ∴OC = 3√2, 即⊙O半径为3√2． 3.(1)见解析; (2)1. (1)证明: 连接OA, 过O作OM ⊥AC于点M, ∵AB = AC且O为BC中点,所以AO平分∠BAC, ∵⊙ O与AB相切于点D, ∴OD ⊥AB, ∵OM ⊥AC,所以OM = OD = r, 所以AC是⊙O的切线. (2)过点O 作 ON ⊥GH, FN =\(1 _{2}\)GF, ∴OM = NH, ∵OD ⊥AB, GH ⊥AC, 所以∠GOF=∠DOB = 90° −∠B, ∠GFO = ∠GFH = 90° −∠C, ∵AB = AC, 所以∠B= ∠C, ∠GOF = ∠GFO, GO = GF, ∵OF = OG, 所以△OGF为等边三 角形, ∴∠GOF = ∠DOE = 60°, ∵OD = OE,所以 △ODE 为等边三角形, ∴OD = DE = r = 2,所以 FN =\(1 _{2}\)GF =\(1 _{2}\)DE = 1, NH = OM =r= 2, ∴FH = NH −FN = 1. 4.C 因为⊙O是\(\Delta{}ABC\)的内切圆, 所以OB, OC分别平分∠ABC, ∠ACB, 因为= 46∘, ∠ACB = 84∘, 所以∠OBC = 1\(1 _{2}\)∠ABC = 23∘, ∠OCB = _{2}\(\times\)∠ACB= 42∘, ∴∠BOC = 180∘−∠OBC −∠OCB = 115∘. 故选: C. 5.(1)见解析; (2)2√3． (1)连接OD,⊙O是△ABC的外接圆, AB是⊙O的直径, ∴∠ACB = 90∘,OA= OD, ∴∠OAD = ∠ODA, ∵DE ∥ BC, ∴∠AED = ∠ACB = 90∘,∵∠BAC的平分线交⊙O于点D, ∴∠EAD = ∠BAD, ∴∠EAD =∠ADO,∴OD ∥ AE, ∴∠ODE = 180∘−∠AED = 90∘, ∵OD为⊙O的半径, ∴DE是⊙O的切线; (2)设OD交BC于点F, ∵∠BAC = 60∘, ∠ACB = 90∘,∴∠ABC = 30∘, ∵CE ∥DF, DE ∥CF, ∠E = 90∘, ∴ 四边形CEDF为矩形,∴∠DFC = 90∘, DF = CE = √3, ∴∠OFB = 90∘, 设⊙O的半径为r, 则:OB= OD = r, OF = OD −DF = r −√3, ∵∠OFB = 90∘, ∠ABC = 30∘, ∴OB =2OF,∴r = 2(r −√3), ∴r = 2√3, ∴⊙ O的半径为2√3． 6.(1)见解析; (2)3√5. (1)证明: 如图连接OC, OA = OC, 所以∠AOC = 2∠ABC = 90°, ∠OAC=∠OCA = 45°, ∵AD是⊙O的切线, ∴ ∠OAD = 90°, ∴∠OAC = ∠EAC=45°, ∵CE = AE, ∴∠ECA = ∠EAC = 45°, 所以∠AEC = 90°, 所以四边形 OAEC为正方形, ∴∠ECO = 90°, ∴EC ⊥OC, ∴CE是⊙O的切线; (2)因为四边形OAEC为正方形, 所以EC ∥AF, 所以△ECD ∼△AFD,EC:AF= DE: AD, 因为DE = 2AE = 4, 所以AF = 3, OF = 1, CF = √5, CD=2√5, DF = CF + CD = 3√5. 7.20∘ 连接OC, 由圆周角定理得, ∠COD = 2∠A = 70∘, 因为CD为⊙O的切线,所以OC ⊥CD, ∴∠D = 90∘−∠COD = 20∘. 8.(1)见解析; (2)①2; ②0或4． (1)∵正六边形ABCDEF内接于⊙O, ∴AB = BC = CD = DE = EF =FA,∠A = ∠ABC = ∠C = ∠D = ∠DEF = ∠F. ∵点P, Q同时分别从A,