25.(10分)
(1)证明:?AB?AC,点D是BC的中点?AD?BC(三线合一).....2??BE?CE(线段的垂直平分线性质)......2?(2)?BF?AC,?BAC?45???ABF??BAF?AF?BF......2??BF?AC,AD?BC??AFE??BFC?90?,?ADC?90???CAD??C?90?,?CBF??C?90???CAD??CBF.......2???AEF??BCF(AAS)?AE?BC..........2?
26.(10分)
解:设时间为t秒,点Q速度为xcm/s(1)全等当t?1时,BP?CQ?1.5?PC?2.5又?D是AB的中点?BD?2.5?BD?PC?AB?AC??B??C??BPD??CQP(SAS).............5?(2)?速度不等?BP?CQ?满足全等必有BP?PC,BD?CQ?1.5t?2,xt?2.5?x?27.(12分)
1515?当x?时,两三角形全等.........5?88
解:(1)GC垂直平分BB?(垂直、平分各1?)..............2?由折叠知:BC?B?C,?BCG??B?CG?CG垂直平分BB?(三线合一)........2?(?点B和点B?关于GC对称?GC垂直平分BB?)(2)由折叠知:BF?CF,EF?BC?BB??CB?........2?又BC?CB???BB?C是等边三角形??BCB??60?.........2?(3)是.............1?由折叠知:GH?CC?,CH?C?H?GC?GC?.......2???BCB??60?,?BCG??B?CG??BCG?30?,又?BCD?90???GCC??60???GCC?是等边三角形..............1?(其它答案参照给分)
29.(12分)
(1)证明:?BD?m,CE?m,??ADB??CEA?90???DBA??DAB?90???BAC?90???DAB??EAC?90???DBA??EAC................2?又?AB?AC??DBA??EAC(AAS)?AD?CE,BD?AE?BD?CE?DE......................2?(2)?BAC??
证明:??ADB??CEA????DBA??DAB?180????BAC????DAB??EAC?180?????DBA??EAC................2?又?AB?AC??DBA??EAC(AAS)?AD?CE,BD?AE?BD?CE?DE......................2?(3)?ADB??AEC?180???,CE?BD?DE...........2?证明:??ADB??CEA?180?????DBA??DAB????BAC????DAB??EAC????DBA??EAC又?AB?AC??DBA??EAC(AAS)?AD?CE,BD?AE?CE?BD?DE......................2?
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