1)已知,如图甲,MN是平行四边形ABCD外的一条直线,AA’、BB’、CC’、DD’都垂直于MN,A’、B’、C’、D
1个回答

如图,过点B'做B'B''∥AB交AA'于点B'',过点C做CC''∥CD交DD'于点C''

∴四边形ABB'B''和四边形CC'C''D是平行四边形,且B'B''∥C'C''

∴B'B''=AB=CD=C'C'',BB'=AB'',CC'=DC'',∠B''B'A'=∠C''C'D'

又∵∠B''A'B'=∠C''D'C'=90º

∴△B''B'A'≌△C''C'D'

∴A'B''=D'C''

∴A'B''+AB''+CC'=D'C''+B'B+C''D

∴AA'+CC'=BB'+DD'

(2)CC'+BB'+DD'=AA'

如图,过点B'做B'B''∥AB交AA'于点B'',过点C做CC''∥CD交D'D的延长线于点C''

∴四边形ABB'B''和四边形CC'C''D是平行四边形,且B'B''∥C'C''

∴B'B''=AB=CD=C'C'',BB'=AB'',CC'=DC'',∠B''B'A'=∠C''C'D'

又∵∠B''A'B'=∠C''D'C'=90º

∴△B''B'A'≌△C''C'D'

∴B''A'=C''D'

∵B''A'=AA'-AB''=AA'-BB',C''D'=C''D+DD'=CC'+DD'

∴AA'-BB'=C''D+DD'=CC'+DD'

∴AA'=BB'+C''D+DD'

(3)AA'=CC'+DD'

如图,过点C做CD''∥CD交D'D的延长线于点D''

∴四边形CC'D''D'是平行四边形,且C'D''∥CD∥AB

∴AB=CD=C'D'',∠ABA'=∠D''C'D',且CC'=DD''

又∵∠AA'B=∠D''D'C'=90º

∴△ABA'≌△D''C'D'

∴AA'=D'D''

∴AA'=DD''+DD'

∴AA'=CC'+DD'

在B、D之间时,AA'+CC'=BB'+DD'

如图,同理可证,△B'B''A''≌△C'D''D'

∴A'B''=D'D''

∴AA'-BB'=CC'-DD'

∴AA'+CC'=BB'+DD'

在A、D之间时与在B,D之间相同

过A点时,CC'=BB'+DD'

如图,过D'点做D'C''∥CD交CC'于点C''

则△ABB'≌△D'C''C'

∴CC''=BB'

∴CC'-CC''=CC'-DD'=BB'

∴CC'=BB'+DD'