证明：AB^2●DC+AC^2●BD-AD^2●BC
=AB^2●DC+AC^2●BD-AD^2●(BD+DC)
=(AB^2-AD^2)×DC+(AC^2-AD^2)×BD
过A点作AE⊥BC于点E
1、若点E与B(或C)重合（自己作图）
则：(AB^2-AD^2)×DC+(AC^2-AD^2)×BD
=(AC^2-AD^2)×BD-(AD^2-AB^2)×DC
=[(AB^2+BC^2)-(AB^2+BD^2)]×BD-BD^2×DC
=(BC^2-BD^2)×BD-BD^2×DC
=(BC+BD)(BC-BD)×BD-BD^2×DC
=(BC+BD)DC×BD-BD^2×DC
=BD×DC×(BC+BD-BD)
=BC×DC×BD
2、若点E在BC的延长线上（自己作图）
则：(AB^2-AD^2)×DC+(AC^2-AD^2)×BD
=[(AE^2+BE^2)-(AE^2+DE^2)]×DC+[(AE^2+CE^2)-(AE^2+DE^2)]×BD
=(BE^2-DE^2)×DC+(CE^2-DE^2)×BD
=(CE+DE)(CE-DE)×BD-(DE+BE)(DE-BE)×DC
=(CE+DE)DC×BD-(DE+BE)BD×DC
=BD×DC×[CE+DE-(DE+BE)]
=BC×DC×BD
3、若点E在BC上（自己作图）
则：(AB^2-AD^2)×DC+(AC^2-AD^2)×BD
=[(AE^2+BE^2)-(AE^2+DE^2)]×DC+[(AE^2+CE^2)-(AE^2+DE^2)]×BD
=(BE^2-DE^2)×DC+(CE^2-DE^2)×BD
=(DE+BE)×BD×DC+(CE-DE)×DC×BD
=BD×DC×(DE+BE+CE-DE)
=BC×DC×BD
综上所述AB^2●DC+AC^2●BD-AD^2●BC=BC●DC●BD