解:延长BP交AC于点E,∵AD为∠BAC的平分线,∴∠BAP=∠EAP,∵BP⊥AD于D,∴∠APB=∠APE=90°,在△APB和△APE中,∵ ∠BAP=∠EAP AP=AP ∠APB=∠APE=90° ,∴△APB≌△APE(ASA),∴AB=AE=8,∵AC=12,∴EC=12-8=4,∵△APB≌△APE,∴BP=EP,∵F是BC的中点,∴PF= 1 2 EC= 1 2 ×4=2.