(1)证明:连接OD,∵AP切半圆于D,∠ODA=∠PED=90°,又∵OD=OE,∴∠ODE=∠OED,∴∠ADE=∠ODE+∠ODA,∠AEP=∠OED+∠PED,∴∠ADE=∠AEP,又∵∠A=∠A,∴△ADE∽△AEP;(2)解:∵△AOD∽△ACB,∴0A CA =OD CB =AD AB ,∵AB=4,BC=3,∴AC=5,∴OD=3 5 OA,AD=4 5 OA,∵△ADE∽△AEP,∴AE AP =AD AE =DE EP ,∵AP=y,OA=x,AE=OE+OA=OD+OA=8 5 OA,∴AE AP =AD AE =4 5 OA 8 5 OA =1 2 ,则y=...