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參考答案評分標準的精神給分.123456BDACDB∴∠AFB=∠DEC······················································································1分∴BE+EF=CF+EF·····································································∴BF=CE················································································∴△ABF≌△DCE······································································∴∠B=∠C······························································································5分∴AB∥CD·························································BC2=132=169··························2+BD2=BC2·············································································即△BCD為直角三角形·······································∴∠ADC=180°-∠CDB=90°··························································································4分AD2+DC2=AC2·················································································5分設AC=x.則(x-5)2+122=x2··································································∴x=16.9··············∴∠BDC=∠BEC=90°··············································································1分∴EM=BC,DM=BC·······························∴EM=DM································································即△DEM為等腰三角形·····································20本題8分)〈∠A=∠A,∴△ABD≌△ACE··········································∴∠ABD=∠ACE·······················································································2分∴∠ABC=∠ACB·······················································································3分∴∠ABC-∠ABD=∠ACB-ACE∴∠OBC=∠OCB(也可證△EBC≌△DCB得到該結論)···························4分∴OB=OC··············∴點O在BC的垂直平分線上··················∴點A在BC的垂直平分線上····························································∴AO⊥BC··········································21本題8分)∴∠BAE+∠EAC=∠CAD+∠EAC∴∠BAC=∠EAD·······················································································2分∴∠AED=∠B=60°···············································································∴∠ADN=∠ACM,DE=BC········································································4分∴AM=AN,∠MAC=∠DAN.············································∴∠MAN=∠MAC+∠CAN=∠DAN+∠CAN=∠DAC=60°··························································································7分∴△AMN是等邊三角形···································22本題8分)∴∠C=∠DBE=90°··················································································〈lAB=DE.∴Rt△ABC≌Rt△EDB(HL)·············································=BD-AC·························································································4分(2)連接AD、AE.APBQCDAPBQCD∴∠AOD=∠BDE+∠OBD=∠ABC+∠OBD=∠EBD=90°··························∴SACBDb+a)a=c2+b(a-b)·············································································7分2+b2=c2································································23本題7分)(1)5·····························································································(2)角平分線正確2分,點D標正確1分··········································24本題8分)(1)二·························································(2)過點A作AM⊥BD,AN⊥CD,分別交BD、CD的延長線于點M、N·············3分NBNBAA DEM∠BDE DEM∴∠ADM=∠AND∴△AMD≌△AND········································∴AM=AN,MD=ND··························(AM=ANlAB=AC.∴Rt△AMB≌Rt△ANC(HL)·················25本題9分)【問題解決】∴△DAE≌△DBC····································∴∠EAC+∠ACB=180°···············································································3分∴∠EAC=180°-∠ACB=90°=∠ACB∴AB=EC····················································【問題再探】∴BD=FD··············
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