Question

如图,在三角形ABC中,AB=AC,点D为底边上的任意一点,试说明

Answer 1

作AE⊥BC,设D在B、E间

因AB=AC,所以BE=BC/2

AB^2=AE^2+BE^2

AD^2=AE^2+DE^2

AB^2-AD^2

=(AE^2+BE^2)-(AE^2+DE^2)

=BE^2-DE^2

=BE^2-(BE-BD)^2

=BE^2-(BE^2-2BE*BD+BD^2)

=2BE*BD-BD^2

=BD*(2BE-BD)

=BD*(BC-DB)

=BD*DC