如图所示, 在 $\triangle \mathrm{ABC}$ 中, $\angle \mathrm{BAC}=90^{\circ}, \mathrm{AB}=\mathrm{AC}$,

$\mathrm{MN}$ 是经过点 A 的直线, $\mathrm{BD} \perp \mathrm{MN}, \mathrm{CE} \perp \mathrm{MN}$, 垂足分别为 D, E.

(1) 求证: (1) $\angle \mathrm{BAD}=\angle \mathrm{ACE}$; (2) $\mathrm{BD}=\mathrm{AE}$.

(2) 请写出 $\mathrm{BD}, \mathrm{CE}, \mathrm{DE}$ 三者间的数量关系式, 并证明.

<ImageHere>
(1) 证明: (1) $\because \angle \mathrm{BAC}=90^{\circ}$,

$\therefore \angle \mathrm{BAD}+\angle \mathrm{CAE}=90^{\circ}$,

$\because \mathrm{CE} \perp \mathrm{MN}, \quad \therefore \angle \mathrm{ACE}+\angle \mathrm{CAE}=90^{\circ}$,

$\therefore \angle \mathrm{BAD}=\angle \mathrm{ACE}$

(2) $\because \mathrm{BD} \perp \mathrm{MN}, \quad \mathrm{CE} \perp \mathrm{MN}$,

$\therefore \angle \mathrm{BDA}=\angle \mathrm{AEC}=90^{\circ}$,

在 $\triangle A B D$ 和 $\triangle C A E$ 中,

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

$\therefore \triangle \mathrm{ABD} \cong \triangle \mathrm{CAE}$,
$\therefore \mathrm{BD}=\mathrm{AE}$

(2) $\mathrm{BD}=\mathrm{CE}+\mathrm{DE}$. 证明如下:

$\because \triangle \mathrm{ABD} \cong \triangle \mathrm{CAE}$

$\therefore \mathrm{BD}=\mathrm{AE}, \mathrm{AD}=\mathrm{CE}$.

$\because \mathrm{AE}=\mathrm{AD}+\mathrm{DE}$,

$\therefore \mathrm{BD}=\mathrm{CE}+\mathrm{DE}$.