(1)证明:∵AB为直径,∴∠ACB=∠ADB=90°
∵BD平分∠ABC∴∠CBF=∠FBA
∵∠DAF+∠AFD=90° ∠CBF+∠BFC=90°
∠AFD=∠BFC(对顶角相等)
∴∠DAF=∠CBF=∠FBA
∵∠FBA+∠DAE=90° ∠EDA+∠DAE=90°
∴∠FBA= ∠EDA
∴∠DAF=∠EDA
∴AP=DP(等角对等边)
∵DE⊥AB
∴∠FBA+∠BDE=90°
∵∠DAF+∠AFD=90° ∠DAF=∠FBA
∴∠BDE=∠AFD
∴DP=PF(等角对等边)
∴AP=PF(等量代换)
(2)∵∠ADF=∠BDA=90° ∠DAF=∠FBA
∴△ADF~△BDA
∴DA/DB=AF/BA
∵O的半径为5∴BA=10
又∵AF=15/2
∴DA/DB=AF/BA=3/4
∵tan∠ABF=DA/DB
∴tan∠ABF=3/4