题目 #20476

过点A作AG ∥CD, 交EF于点H, 交BC于点G, ∵AD ∥EF ∥BC,所以四边形AHFD, AGCD均为平行四边形, ∴ CG = HF = AD = 3,因为EF = 5, ∴EH = EF −HF = 5 −3 = 2, ∵EF ∥BC, ∴△AEH ∼△ABG,所以BG\(AB _{EH}\)= _{AE}, ∵ AE: EB = 3: 2, ∴ _{EH}= _{AE}= _{3}, ∴BG = _{3}EH = _{3},所以BC = BG + CG = _{3}+ 3 = _{3}, 故答案为 _{3}.^{19} BG AB 5 5 10 10 19 8.见解析. ∵∠BCE = ∠ACD, ∴∠BCE + ∠ACE = ∠ACD + ∠ACE, ∴∠DCE =∠ACB, 又 CD ⋅BC = CE ⋅AC,则 CD\(CE _{CA}\)= _{CB}, ∴△ ABC ∼△DEC. 9.(1)见解析; (2) _{2}.^{9} (1)在Rt △ABC中, ∠ACB = 90∘, ∵CD是△ABC的中线,∴AD = BD = CD, ∴∠DCB = ∠DBC, ∵AE ⊥ CD,∴∠AEC = ∠ACB = 90∘, ∴∠CAE = 90∘−∠ACE = ∠DCB,∴∠CAE = ∠DBC, ∴△ACE ∼△BAC; (2)∵AC = 3, CE = 1,由(1)知 △ACE ∼△BAC, ∴ CE AC 1\(3 _{AC}\)= _{AB}, ∴ _{3}=_{AB}, ∴AB = 9, ∵CD是△ABC的中线, ∴CD =\(1 _{2}\)AB = _{2}.^{9} 10.2: 3 因为DE ∥BC, ∴△ADE ∼△ABC, ∴ DE AD AD\(2 _{BC}\)= _{AB}, ∵AD = 2BD, ∴ _{AB}= _{3},所以DE: BC = 2: 3, 故答案为2: 3. 11.(1)见详解; (2)CD = 4. (1)∵CA = CB, ∴∠A = ∠B, ∵∠CDB = 2∠B, ∴∠CDB = 2∠A, 又∵∠CDB= ∠A + ∠ACD, ∴∠ACD = ∠B, ∵∠A = ∠A, ∴△ABC ∼△ACD. (2)∵AC = 6, ∴BC = AC = 6, ∵△ABC ∼△ACD, ∴ AB BC 9\(6 _{AC}\)= _{CD}, ∴ _{6}= _{CD},解得CD = 4. 12.1: 2 ∵四边形 ABCD 是平行四边形, ∴BC = AD, AD ∥BC, ∵ 点 E 是 AD 的中点,∴DE =\(1 _{2}\)AD, ∴DE: BC = 1: 2, ∵ AD ∥BC, ∴△DEF ∼△BCF,∴C△DEF: C△BCF= 1: 2, 故答案为1: 2． 13.8 DE AD 1 4 1 ∵DE ∥BC, ∴△ADE ∼△ABC, ∴ _{BC}= _{AB}= _{2}, 即 _{BC}= _{2}, ∴BC = 8cm, 故答案为8. 14.见解析. ∵△ACE ≅△ABF, ∴∠CAE = ∠QAF, AE = AF, ∵ AE AC AE\(AC _{AQ}\)= _{AE},∴_{AQ}= _{AF}, ∴△ACE ∼△AFQ. 5