$解：?(1)?∵?AD//BC?$
$∴只要?AP = BQ，?四邊形?ABQP ?是平行四邊形$
$設(shè)?xs ?后四邊形?ABQP ?是平行四邊形$
$則?AP= x，??BQ=6-2x?$
$由?x=6-2x?$
$解得?x=2?$
$∴?2s ?后四邊形?ABQP ?是平行四邊形$
$?(2)?同理，只要?DP = OQ，?四邊形?PQCD?是平行四邊形$
$設(shè)?xs ?后四邊形?PQCD?是平行四邊形$
$則?DP=8-x，??CQ=2x?$
$由?8-x= 2x?$
$解得?x=\frac {8}{3}?$
$∴?\frac {8}{3}s ?后四邊形?PQCD?是平行四邊形$
