(1)證明:連接$CE,$因?yàn)?D$是$BC$的中點(diǎn),且$DE\perp BC,$所以$CE = BE。$又因?yàn)?BE^{2}-EA^{2}=AC^{2},$所以$CE^{2}-EA^{2}=AC^{2},$即$EA^{2}+AC^{2}=CE^{2},$根據(jù)勾股定理的逆定理可知$\triangle ACE$是直角三角形,所以$\angle A = 90^{\circ}。$
(2)解:因?yàn)?DE = 3,$$BD = 4,$所以$BE^{2}=DE^{2}+BD^{2}=3^{2}+4^{2}=25 = 5^{2},$所以$BE = CE = 5。$因?yàn)?BC = 2BD = 8,$在$Rt\triangle BAC$中,由勾股定理得$BC^{2}-BA^{2}=AC^{2},$即$64-(5 + AE)^{2}=25 - AE^{2},$
$\begin{aligned}64-(25 + 10AE+AE^{2})&=25 - AE^{2}\\64 - 25 - 10AE - AE^{2}&=25 - AE^{2}\\39 - 10AE - AE^{2}+AE^{2}&=25\\-10AE&=25 - 39\\-10AE&=-14\\AE&=\frac{7}{5}\end{aligned}$