如图：四边形ABCD是等腰梯形或矩形．证明如下：∵PA=PB=PC=PD，AB=CD，∴△PAB≌△PDC，∠PAB=∠PBA=∠PCD=∠PDC．又∵∠PDA=∠PAD，∴∠BAD=∠CDA．同理∠ABC=∠DCB．于是∠BAD+∠ABC=12×360°=180°，∴AD∥BC．...