证明： ∵ \(EF \bot DE\)， \(DG⊥DE，FG⊥EF\)，
\(∴∠DEF=∠EDG=∠EFG=90°\)
∴四边形\(DEFG\)是矩形
如图，作\(EM \bot BC\)于\(M\)，\(EN \bot CD\)于\(N\)，
\(\therefore \angle MEN = 90^\circ \)，
\(\because \)点\(E\)是正方形\(ABCD\)对角线上的点，
\(\therefore EM = EN\)，
\(\because \angle DEF = 90^\circ \)，
\(\therefore \angle DEN = \angle MEF = 90^\circ - \angle FEN\)，
\(\because \angle DNE = \angle FME = 90^\circ \)，
在\(\Delta DEN\)和\(\Delta FEM\)中，
\(\left\{ {\begin{array}{*{20}{l}} {\angle DNE = \angle FME} \\ {EN = EM} \\ {\angle DEN = \angle FEM} \end{array}} \right.\)，
\(\therefore \Delta DEN \cong \Delta FEM(ASA)\)，
\(\therefore EF = DE\)，
\(\because \)四边形\(DEFG\)是矩形，
\(\therefore \)矩形\(DEFG\)是正方形。
