19．（1）证明∵PA=AB=2a，PB=2a,

∴PA²+AB²=PB²，∴∠PAB=90°，即PA⊥AB．

同理PA⊥AE．                                                                                   3分

∵AB∩AE=A，∴PA⊥平面ABCDE．                                           5分