过点C作CD⊥CP,使CD=CP
∠PCD=∠PCB+∠BCD=90°
∠ACP+∠PCB=∠ACB=90°
∠ACP=∠BCD
AC=BC
CP=CD
△ACP≌△BCD
AP=BD=6
CD=CP=4
∠PCD=90°
PD=4√2,∠CPD=45°
在三角形BCD中,
BD^2=36
PB^2=4,PD^2=(4√2)^2=32
BD^2=PB^2+PD^2
∠BPD=90°
∠CPB=∠CPD+∠BPD=45°+90°=135°