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八年级上册数学补充习题答案(2020)

时间:2020-07-25 00:24:32

  1.△ACB ≌ NMR,△DEF ≌ △QOP.
2.在△ABC和△CDA中,
∵AB = CD, ∠BAC= ∠DCA,
AC = CA,
∴△ABC ≌ △CDA(SAS).
3.∵AB ⊥ CD,∠ABC = ∠DBE = 90°.又
AB = DB,BC = BE,
∴△ABC ≌△DBE(SAS).
4.(1) ∵AD = AE, ∠1 = ∠2, AO = AO,
∴△AOD ≌ △AOE( SAS).
(2) ∵AC = AB,∠1 = ∠2, AO = AO,
∴△AOC ≌ △AOB( SAS).
(3) ∵AB = AC,∠BAD = ∠CAE,AD = AE,∴△ABD ≌△ACE( SAS).
1.∵ AD是△ABC的中线,
∴ BD = CD.又∠BDN = ∠CDM,
DN = DM,
∴ △BDN ≌ △CDM( SAS).
2.∵ AD是△ABC的中线,
∴BD = CD.
∵ AD ⊥ BC,
∴∠ADB = ∠ADC = 90°.在△ABD和
△ACD中,
∵AD = AD,∠ADB = ∠ADC, BD = CD,
∴△ABD ≌ △ACD(SAS).
∴ AB = AC.
3.在△ABC和△DEF中,
∵AB = DE, ∠B = ∠E, BC = EF,
∴△ABC ≌ △DEF(SAS).
∴ ∠ACB = ∠DFE.
∵∠ACF + ∠ACB = ∠DFC + ∠DFE = 180°,
∴ ∠ACF = ∠DFC.
∴ AC ∥ DF.
4.(1) 利用(SAS)证明;
(2) 共可画14条.1.∵ AB ∥ DC,AD ∥ BC,
∴ ∠BAC = ∠DCA,∠BCA = ∠DAC.
在△ABC和△CDA中,
∵∠BAC = ∠DCA,AC = CA,
∠BCA = ∠DAC,
∴ △ABC ≌ △CDA(ASA). ∴ AB = DC,
AD = BC.
2.在△ABE和△ACD中,
∵∠A = ∠A,AB = AC,∠B = ∠C,
∴ △ABE ≌ △ACD(ASA).
∴ AD = AE.
∴ AB - AD = AC - AE.即DB = EC.
3.∵ ∠3 + ∠AOB = ∠4 + ∠AOC = 180°,∠3 = ∠4,
∴∠AOB = ∠AOC.在△AOB和△AOC中,
∵ ∠1 = ∠2, AO = AO,∠AOB = ∠AOC,
∴ △AOB ≌ △AOC(ASA).
∴ OB = OC.

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