题目 #18519

如图, 延长AB交DE于T, ∵BC ∥DE, ∴∠ABC = ∠ATE = 38∘, ∵AT ∥EC,所以∠CER = ∠ATE = 38∘, ∴ 1 ∠DEB = ∠CEB = _{2}(180∘−38∘) = 71∘,所以∠DEF = 180∘−∠DEB = 109∘, 故答案为109∘. 2.B 因为AB的垂直平分线交BC于点D，∴AD = BD, ∴△ACD的周长为AC+CD+AD = AC + CD + BD = AC + BC = 6 + 8 = 14, 故选B. 3.(1)证明见解析(2)67.5∘ (1)因为BE ⊥AC, AD ⊥BC, ∴∠AEB = ∠ADC = ∠BDF = 90∘, ∵∠AFE等于∠BFD, ∠FBD + ∠BFD = 90∘, ∠AFE + ∠DAC = 90∘, ∴∠ACD = ∠BFD.又∵∠BDF = ∠ADC, BD = AD, ∴△BDF ≅△ADC所以BF = AC (2)因为AD ⊥BC, AD = BD, ∴△ABD是等腰直角三角形, ∴∠ABD = ∠BAD等于45∘, ∵AB = BC, ∴∠BAC = 1 ∠BCA = _{2}(180∘−45∘) = 67.5∘, ∴∠FAE等于∠BAC −∠BAD = 67.5∘−45∘= 22.5∘, ∴∠AFE = 90∘− ∠FAE = 67.5∘. 4.64∘ 因为一张长方形纸片ABCD沿EF折叠后, 点C, D分别落在C′, D′的位置上，∴AD ∥BC, ∠CEF = ∠FEG, ∴ ∠FEC = ∠FEG = ∠EFG, ∵∠EFG=58°,∴∠FEC = ∠FEG = 58∘, ∴∠BEG = 180∘−∠FEG −∠FEC = 180∘−58°−58∘= 64∘. 故答案为：64°. 5.C. 5,6,10 3 + 5 = 8, A错误；4 + 5 = 9，B错误；5 + 6 = 11 > 10, C正确；6 + 7 = 13 < 14, D错误. 6.证明见解析 因为四边形ABCD是矩形, ∴AB = CD, ∠ABC = ∠BCD = 90∘, ∵BF = CE,所以BF + BC = CE + BC, ∴BE = CF, 在△ABE和△DCF中, ∵AB = CD,且∠ABC = ∠BCD = 90∘, BE = CF, ∴△ABE ≅△DCF,所以∠AEB = ∠DFC, ∴ GE = GF. 7.28∘ 设∠B = x, ∠C = y, ∵DA = DC, ∴∠DAC = ∠C = y, ∵∠ADB = ∠C + ∠DAC = 2y,因为BA = BD, ∴∠ADB = ∠BAD = 2y, ∴∠BAD + ∠DAC = 2y + y = 3y = 114∘,所以y = 38∘= ∠C, ∴∠B = 180∘−114∘−38∘= 28∘. 故答案为28∘. 8.11 因为AB ∥DE, ∴∠B = ∠E, ∵AC = DF, ∠A = ∠D, ∴△ABC ≌△DEF, ∴BC = EF, BC −FC=EF −FC, BF = CE, ∵ BF = 4, ∴CE = 4, ∴BE = 4 + 3 + 4 = 11. 9.(1)证明见解析(2)58∘ (1)因为BE = CD, ∴BE −DE = CD −DE, ∴BD = CE. 在△ABD和△ACE中, ∵AB = AC, BD = CE, AD = AE, ∴ △ABD ≅△ACE. ∴BD = CE (2)因为△ABD ≅△ACE, ∴∠B = ∠C = 36∘, ∠BAD = ∠CAE = 25∘, ∴∠BAC等于180∘−∠B −∠C = 108∘, ∴ ∠DAE = ∠BAC −∠BAD −∠CAE = 108∘−25∘−25∘= 58∘. 10.48∘ 设∠1 = x, 则∠3 = ∠4 = ∠1 + ∠2 = 2∠1 = 2x, 在△ABC中, ∠2 + ∠4 + ∠BAC = 180∘,即x + 2x + 81 = 180, 解得x = 33, ∴∠CAD = ∠BAC −∠1 = 81∘−33∘= 48∘, 故答案为48∘. 11.5.5 因为AC = CD，AC = 3.5，∴CD = 3.5，∵BD = 9，∴BC = BD −CD = 9 −3.5 = 5.5,因为EC ⊥BD，∴ ∠ACB = ∠DCE = 90°，在Rt △ACB和Rt △DCE中，{AB = DE AC = DC,所以Rt △ACB ≌Rt △DCE(HL), ∴CE = CB = 5.5. 12.(1)115∘. (2)证明见解析 (1)因为AB是⊙O的直径，∴∠ADB = ∠ACB = 90∘, 又∠ABC = 25∘, ∴∠CAB等于90∘−25∘= 65∘, ∵ 四边形ABEC是⊙O内接四边形, ∴∠CEB + ∠CAB = 180∘所以∠CEB = 180∘−∠CAB = 115∘ ⏜ ⏜ (2)连接 AI, ∵点 I 为△ABC的内心, ∴∠CAI = ∠BAI,所以∠ACI = ∠BCI =\(1_{2}\)∠ACB = 45∘, ∴AD = BD, ∴ ∠DAB = ∠DCB = ∠ACI, AD = BD,因为∠DAI = ∠DAB + ∠BAI, ∠DIA = ∠ACI + ∠CAI所以∠DAI = ∠DIA, ∴ DI = DA. 13.32 因为AB = AD, AD = DC, ∴∠ABD = ∠ADB, ∠DAC = ∠DCA, ∵∠C = 37∘,所以∠DAC = 37∘, ∴∠ADB = ∠DAC + ∠C = 37∘+ 37∘= 74∘. ∴∠ABD = 74∘,所以∠BAD = 180∘−74∘−74∘= 32∘. 故答案为32. 14.13 因为DE是AB的垂直平分线，∴AE = BE,因为\(\Delta{}BCE\)的周长为8, 所以AB + BC = 8因为AB −BC = 2, 所以AB = 5, BC = 3,因为AB = AC, 所以AC = 5所以△ABC的周长是AC + AB + BC = 5 + 5 + 3 = 13. 15.45° 因为∠B = 40°，∠C = 50°，∴∠BAC = 180° −∠B −∠C = 90°, ∵AD平分∠BAC,∴∠BAD = ∠CAD =\(1 _{2}\)∠BAC = 45°, ∵DE ∥AB, ∴∠ADE = ∠BAD =45°，故答案为45°.