这题确实不好作,题目也没告诉四边形是平面四边形：
|AB|=|AD|,|CB|=|CD|
即：△ADC≌△ABC,即：∠ADC=∠ABC
即：AD·DC=AB·BC
AC·DB=(AD+DC)·(AB-AD)=AB·AD+AB·DC-|AD|^2-AD·DC
=AD·(AB-AD)+AB·DC-AB·BC
=AD·(AB-AD)+AB·(DC-BC)
=AD·DB+AB·(CB-CD)
=AD·DB+AB·DB=(AD+AB)·DB
取DB边中点E,则：AC·DB=2AE·DB
△ABD是等腰三角形,故：AE⊥DB
即：AE·DB=0,故：AC·DB=0
即：AC⊥DB