證明：如圖（1）過點C作PE的垂線交PE于K,則,∠CKE=90°
       ∵ PE⊥AB  CD⊥AB
       ∴  ∠CDE=∠DEK=∠CKE=90°
      ∴ DCKE為矩形
       ∴ EK=CD   CK‖AB
     ∴ ∠ABC=∠KCP (兩直線平行,同位角相等)
    又∵ AB=AC
      ∴ ∠ABC=∠ACB  
   又∵ ∠ACB=∠FCP(對頂角相等)
       ∴ ∠ABC=∠FCP
     ∴ ∠KCP=∠FCP
 在Rt△KCP和Rt△FCP中
 ∵ ∠PKC=∠PFC=90°
    ∠KCP=∠FCP
      PC=PC
   ∴ Rt△KCP≡Rt△FCP (AAS)
     ∴ PF=PK
     又∵PK=PE-EK=PE-CD
     ∴PF=PE-CD
   ∴ PE=PF+CD
   ∴ PE-PF=CD
                  證畢!