证明:作AE⊥BC于E,则 AB 2 =AE 2 +BE 2 ,AD 2 =AE 2 +DE 2 , 则AB 2 -AD 2 =(AE 2 +BE 2 )-(AE 2 +DE 2 )=+BE 2 )-(AE 2 +DE 2 )=(BE+DE)(BE-DE)=BD•DC, 则AD 2 +BD•DC=AB 2 .