解:∵四边形ABCD是正方形∴OB=OC=1/2AC=5(O 为AC,BD的交点)连接OP则S△OBC=S△OBP+S△OCP∴1/2*OC*OB=1/2*OB*PE+1/2OC*PF∵OB=OC∴OB=PE+PF∴PE+PF=OB=5