解:过点P作PM⊥AB于M,PN⊥CD于N,PG⊥BC于G∵AB∥CD∴∠ABC+∠DCB=90∵PM⊥AB,PG⊥BC,PM=PG∴BP平分∠ABC∴∠CBP=∠ABC/2∵PN⊥CD,PG⊥BC,PN=PG∴CP平分∠DCB∴∠BCP=∠DCB/2∴∠CBP+∠BCP=(∠ABC+∠DCB)/2=90∴∠BPC=180-(∠CBP+∠BCP)=90°
四边形ABCD为正方形时,才能满足题设条件故∠BPC=90°
90°