* BD ∠abc∴∠cbf=∠fba.

∠DAF+∠AFD=90 ∠CBF+∠BFC=90

∠AFD=∠BFC (equal vertex angles)

∴∠DAF=∠CBF=∠FBA

∠∠FBA+∠DAE = 90∠EDA+∠DAE = 90

∴∠fba=∠ Ada

∴∠DAF=∠EDA

∴AP=DP (equilateral)

∵DE⊥AB

∴∠FBA+∠BDE=90

∠∠DAF+∠AFD = 90∠DAF =∠FBA

∴∠BDE=∠AFD

∴DP=PF (equilateral)

∴AP=PF (equivalent substitution)

(2)∫∠ADF =∠BDA = 90∠DAF =∠FBA

∴△ADF~△BDA

∴DA/DB=AF/BA

The radius of ∵O is 5∴BA= 10.

∫AF = 15/2

∴DA/DB=AF/BA=3/4

∫tan∠ABF = DA/DB

∴tan∠ABF=3/4