∵∠A=∠C，∠B=∠D，∴△ABP∽△CPD，∴PD/PB=AC/BD=3/4连结BD，∵AB是直径，∴∠ADB=90°，∴cos∠BPD=PD/PB=3/4，设PD=3X，则PB=4X，由勾股定理得BD=√7X，∴tan∠BPD=BD/PD=√7/3
