直线l与平行四边形ABCD一组对边相交,AA'⊥l,BB'⊥l,CC'⊥l,DD'⊥l,A'、B'、C'、D’为垂足求证：AA'-CC'=BB'-DD'

直线l与平行四边形ABCD一组对边相交,AA'⊥l,BB'⊥l,CC'⊥l,DD'⊥l,A'、B'、C'、D’为垂足求证：AA'-CC'=BB'-DD'
直线l与平行四边形ABCD一组对边相交,AA'⊥l,BB'⊥l,CC'⊥l,DD'⊥l,A'、B'、C'、D’为垂足
求证：AA'-CC'=BB'-DD'

直线l与平行四边形ABCD一组对边相交,AA'⊥l,BB'⊥l,CC'⊥l,DD'⊥l,A'、B'、C'、D’为垂足求证：AA'-CC'=BB'-DD'
证明：作A'E//AB交BB'于E点,作C'F//CD交DD'于F点
易知ABEA'是平行四边形,即
AB=A'E,AA'=BE,A'E//AB
BB'-AA'=BB'-BE=EB'
同理CD=C'F,CC'=DF,DD'-CC'=D'F,CD//C'F
又AB=CD,AB//CD
所以A'E=C'F,A'E//C'F
所以∠EA'B'=∠FC'D'
所以RT△A'B'E≌RT△C'D'F
即B'E=D'F
BB'-AA'=DD'-CC'
AA'-CC'=BB'-DD'