作AE⊥BE,根据勾股定理,AB^2=BE^2+AE^2,
AB^2=(BD+DE)^2+AE^2=BD^2+DE^2+2BD*DE+AE^2.(1),
AC^2=CE^2+AE^2=(CD-DE)^2+AE^2=CD^2-2CD*DE+DE^2+AE^2.(2),
BD=CD,
(1)式+(2)式,
AB^2+AC^2=BD^2+CD^2-2BD*DE+2BD*DE+2AE^2+2DE^2
=BD^2+CD^2+2AD^2=2BD^2+2AD^2.
∴AB^2+AC^2=2BD^2+2AD^2.