如图, 已知 $\triangle A B C$ 中, $\angle C=90^{\circ}, D E \perp A B$ 于 $E, A C=A E$. 求证: $C D=D E$.

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证明: $\because \mathrm{DE} \perp \mathrm{AB}$,

$\therefore \angle \mathrm{AED}=90^{\circ}$

$\because \angle \mathrm{C}=90^{\circ}$

在 Rt $\triangle \mathrm{ACD}$ 和 Rt $\triangle \mathrm{AED}$ 中, $\angle \mathrm{C}=\angle \mathrm{AED}=90^{\circ}$

$\left\{\begin{array}{l}A C=A E \\ A D=A D\end{array}\right.$

$\therefore \mathrm{Rt} \triangle \mathrm{ACD} \cong \mathrm{Rt} \triangle \mathrm{AED}(\mathrm{HL})$,

$\therefore \mathrm{CD}=\mathrm{DE} ;$