\[По\ теореме\ Пифагора:\]
\[AB^{2} + AD^{2} = BD^{2}\]
\[4^{2} + 4^{2} = BD^{2}\]
\[16 + 16 = BD^{2}\]
\[32 = BD^{2}\]
\[BD^{2} + BC^{2} = DC^{2}\]
\[32 + 4^{2} = DC^{2}\]
\[32 + 16 = DC^{2}\]
\[48 = DC^{2} \Longrightarrow DC = \sqrt{48} = 4\sqrt{3}\ (см).\]
\[Ответ:\ \ DC = 4\sqrt{3}\ см.\]