过A作BC垂线交BC于D,由直角三角形的勾股定理得:
AD^2=AP^2-DP^2
AD^2=AC^2-CD^2
所以AP^2-DP^2=AC^2-CD^2
又AB=AC,所以D为BC的中点,那么CD=BC/2,DP=BP-BC/2代入得:
AP^2-(BP-BC/2)^2=AC^2-(BC/2)^2
化简得:AP^2-BP^2+BP*BC=AC^2,又AC=AB:
AP^2-BP^2+BP*BC=AB^2
AP^2=AB^2+BP^2-BP*BC
AP^2=AB^2+BP*(BP-BC)
AP^2=AB^2+BP*CP