人教版新课标八年级上册数学上册期中考试试卷及参考答案

……………………………………………5 分Р(2)由ΔACD ≅ΔCBE 可得CD = BE = 3,所以CE = 3 + 5 = 8,又由AD = CE ,Р可得AD = 8. ……………………………………………8分Р22.(1)∵ΔABC 是等边三角形,∴∠ACB = 60°,AC = BC.Р∵∠F = 30°,∴∠CAF = 60° − 30° = 30°.Р∴∠CAF = ∠F ,∴CF = AC.∴CF = AC = EC.Р∴EF = 2BC .……………………………………………4 分Р(2)成立………………………………………1 分Р先用(1)中证CF = AC 的方法证CH =CF ,∵EF = 2BC,∴BE +CF = BC.Р∵AH +CH = AC,AC = BC,Р∴AH = BE.………………………………………9分Р23.(1)在DC 上截取DM = BD,连结AM .Р先证ΔABD ≅ΔAMD ,得AB = AM ,∠B = ∠AMB.Р∵∠AMD = ∠MAC + ∠C, ∠B = 2∠C,Р∴∠C = ∠MACР∴AM =MCР∴MC = AB.则AB + BD = DC.………………………………………5分Р(2) AB + BD = AC. …………………………………………6 分Р方法一:如图3a 在AC 上截取AM = AB,连结DM .РB CРAРD MР第6 页共6 页作者:树树数数Р先证ΔABD ≅ΔAMD ,可得∠B = ∠AMD.Р再证DM = MC,则MC = BD.…………………………………………………10 分Р方法二:如图3b 延长AB 到M ,使BM = BD ,连结MD.Р∠ABD = ∠M + ∠BDM = 2∠M.Р由∠ABD = 2∠C ,得∠M = ∠C .再证ΔAMD ≅ΔACD.………………………10 分