答案：1.(D)2.(D)3.(A)4.(B)5.(A)6.(A)7.(B)8.(B)9.(C)10.(C)11(B)12.(B)13.4914.2∶5.

15.8m.16.85517.解:設(shè)x3=y4=z∴x+y-zx-18.(1)證明:∵∠BCE=∠ACD,∴∠BCE+∠ACE=∠ACD+∠ACE.∴∠ACB=∠DCE.∵∠A=∠D,∴△ABC∽△DEC.(2)解:∵△ABC∽△DEC,∴S△ABCS△DEC=(CB∴CBCE=2∵BC=6,∴CE=9.19.解:(1)△A1B1C1如圖所示,其中A1的坐標為(0,1).(2)符合條件△A2B2C2有兩個,如圖所示.20.(1)證明:∵四邊形ABCD為菱形,∴CD∥AB,AB=CB.∴∠ACD=∠CAB,∠CAB=∠ACB.∵∠ACD=∠ABE,∴∠ACD=∠ABE=∠CAB=∠ACB.∴△ABC∽△AEB.(2)解:∵△ABC∽△AEB,∴ABAE=AC即6AE=4∴AE=9.21.解:如圖所示,作EH⊥AB于點H,交CD于點G,∵CD⊥FB,AB⊥FB,∴CD∥AB.∴△CGE∽△AHE.∴CGAH=EG即CD-EFAH∴3-1.6解得AH=11.9m.∴AB=AH+HB=AH+EF=11.9+1.6=13.5(m).故旗桿AB的高度為13.5m.22.證明:(1)∵∠ADE=∠B,∴DE∥BC.∴ADAB=AE(2)∵AD2=AF·AB,∴ADAB=AF由(1),得ADAB=AE∴AEAC=AF∵∠A=∠A,∴△AEF∽△ACD.23.(1)證明:∵在矩形ABCD中,DE⊥AM于點E,∴∠B=90°,∠BAD=90°,∠DEA=90°.∴∠BAM+∠EAD=90°,∠EDA+∠EAD=90°.∴∠BAM=∠EDA.在△ADE和△MAB中,∵∠AED=∠B,∠EDA=∠BAM,∴△ADE∽△MAB.(2)解:∵在矩形ABCD中,AB=2,BC=3,M是BC的中點,∴BM=32∴AM=22+(3由(1)知,△ADE∽△MAB,∴AMDA=AB∴523=解得DE=12524.(1)證明:連接OD,如圖所示,∵CD平分∠ACB,∴AD=BD.∴∠AOD=∠BOD=90°.∵DF是☉O的切線,∴∠ODF=90°.∴∠ODF=∠BOD.∴DF∥AB.(2)解:過C作CM⊥AB于點M,如圖所示,∵AB是直徑,∴∠ACB=90°.∴AB=AC2+B∴12AB·CM=12AC∴CM=2.∴BM=BC2-∴OM=OB-BM=52-1=3∵DF∥AB,∴∠OFD=∠COM.又∵∠ODF=∠CMO=90°,∴△DOF∽△MCO.∴CMOD=OMFD,即252=解得FD=15825.解:由題意,得AP=4tcm,CQ=2tcm,則CP=(20-4t)cm.(1)當t=3s時,CP=20-4t=8(cm),CQ=2t=6(cm),由勾股定理,得PQ=CP即當t=3s時,P,Q兩點之間的距離為10cm.(2)Rt△CPQ的面積S=12CP·CQ=12×(20-4t)×2t=20t-4t即S關(guān)于t的函數(shù)解析式為S=20t-4t2(0<t<5).(3)分兩種情況:①當Rt△CPQ∽Rt△CAB時,CP∶CA=CQ∶CB,即(20-4t)∶20=2t∶15,解得t=3;②當Rt