参考答案一、选择题:BBCCDCBAAC二、填空题:111213141516171819206a()2-2yxm100°82532870°或20°13120421、(1)85yx..............4分(2)21012-yxy+..............4分22、化简:12+x..............6分求值:5..............2分23、(1)图略..............6分(2)()1-,1-'A..............2分24、
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:过点A作AH⊥BC于H∵AB=AC∴BH=CH............2分∵AD=AE∴DH=EH............2分∵BD=BH-DH,CE=CH-EH............2分∴BD=CE............2分25、(1)证明;在△ABC中∵AB=AC,点D是BC的中点∴AD⊥BC,BD=CD............3分∴BE=CE............3分(2)△ABE、△ACE、△BCE、△ABC...........4分26、(1)∵AD⊥BC,D为BC中点∴AB=AC∴∠C=∠B...........1分∵∠BAC=2∠B,∠B+∠BAC+∠C=180°............1分∴∠B+2∠B+∠B=180°∴∠B=45°............1分(2)在DH上取一点N使HN=HF∵DH=CF+HF,DH=DN+HN∴CF=DN............1分∵CH⊥DF,HN=HF∴CN=CF∴∠F=∠CNF∵CN=CF,CF=DN∴CN=DN............1分∴∠FDC=∠NCD∵∠CNF=∠FDC+∠NCD∴∠F=2∠FDC............1分(2)由(2)知∠F=2∠FDC,设∠FDC=α,则∠F=2α∵∠BPD=∠F∴∠BPD=2α............1分连接PC交DF于K,∵AD⊥BC,D为BC中点∴BP=CP,∠PCD=∠PBD∵∠BPD=2α∴∠PCD=∠PBD=90°-2α∴∠PKD=∠PCD+∠FDC=90°-α∵AD⊥BC∴∠ADF=90°-∠FDC=90°-α∴∠PKD=∠ADF∴PK=PD............1分过点C作CM⊥EG于M由EF沿着EC折叠可知∠FEC=∠GEC∴CM=CH由(1)知∠ABC=45°,AD⊥BC∴∠BAD=45°∵∠BAC=2∠ABC∴∠DAC=45°∴∠AED=45°+α∴∠FEC=∠CEG=∠AED=45°+α∴∠HEG=90°+2α∴∠DEG=90°-2α∴∠EGC=90°-α∵∠EKC=∠PKD=90°-α∴∠EGC=∠EKC又∵∠GMC=∠KHC=90°∴△GMC≌△KHC............1分由BP:PD=12:5,设BP=12x,PD=5x∴GC=CK=CP-PK=BP-PK=12x-5x=7x∵GC-PD=3∴7x-5x=3∴x=1.5∴GC=7x=10.5............1分27、(1)B(2,0)............2分(2)当0≤t<2时,............1分S=21×2×(2-t)=2-t............1分当t>2时,S=21×2×(t-2)=t-2............1分(3)∵C(-2,4),△ABC是等腰直角三角形∴AB=8由(1)知B(2,0)∴OB=2,OA=6延长AC交y轴于H,连接FD∵△ABC是等腰直角三角形∴∠ACB=90°∴∠CAB=45°∵∠AOH=90°∴∠CHE=∠CAB=45°∴OH=OA=6∵∠ACB=90°∴∠DCH=90°∵∠CHE=45°∴∠CDH=∠CHE=45°∴CH=CD.........1分∵CF⊥CE∴∠DCF+∠ECD=90°∵∠ACB=90°∴∠HCE+∠ECD=90°∴∠HCE=∠DCF又∵CF=CE∴△HCE≌△DCF..........1分∴HE=FD=6-t,∠CDF=∠CHE=45°∵∠CBA=45°∴∠CDF=∠CBA∴FD∥AB∴∠FDM=∠NAM∵M是AD中点∴DM=AM又∵∠FMD=∠NMA∴△DMF≌△AMN∴AN=FD=6-t连接AF、FO∵DM=AM∴AMFΔDMFΔSS=............1分∵△DMF≌△AMN∴AMNΔDMFΔSS=∴AMNΔNFAΔ2SS=∵AMNΔNFOΔ10SS=∴NFAΔNFOΔ5SS=∴5AN=ON∵OA=6∴AN=1∴AN=6-t=1∴t=5............1分∴S=t-2=5-2=3............1分H