过C作CF⊥AB交AB于F.
∵CF⊥AB、ED⊥AB,∴CF∥ED,又BE=CE,∴BD=FD.
∵AC⊥BC、CF⊥AB,∴由射影定理,有:AF×AB=AC^2,
∴(AD-FD)(AD+BD)=AC^2,而BD=FD,∴(AD-BD)(AD+BD)=AC^2,
∴AD^2-BD^2=AC^2.
过C作CF⊥AB交AB于F.
∵CF⊥AB、ED⊥AB,∴CF∥ED,又BE=CE,∴BD=FD.
∵AC⊥BC、CF⊥AB,∴由射影定理,有:AF×AB=AC^2,
∴(AD-FD)(AD+BD)=AC^2,而BD=FD,∴(AD-BD)(AD+BD)=AC^2,
∴AD^2-BD^2=AC^2.