(1)问题发现:

(2)解决问题:

【答案】 (1) 证明: $\because \triangle \mathrm{ABC}$ 和 $\triangle \mathrm{ADE}$ 是顶角相等的等腰三角形,
\begin{aligned} & \therefore A B=A C, A D=A E, \angle B A C=\angle D A E \\ & \therefore \angle \mathrm{BAC}-\angle \mathrm{CAD}=\angle \mathrm{DAE}-\angle \mathrm{CAD} \text {, } \\ & \therefore \angle \mathrm{BAD}=\angle \mathrm{CA} \text { E } \text {. } \\ & \text { 在 } \triangle B A D \text { 和 } \triangle C A E \text { 中 } \\ & \left\{\begin{aligned} \mathrm{AB} & =\mathrm{AC} \\ \angle \mathrm{BAD} & =\angle \mathrm{CAE} \\ \mathrm{AD} & =\mathrm{AE} \end{aligned}\right. \\ & \therefore \triangle B A D \cong \triangle \mathrm{CAE} \text { (SAS), } \\ & \therefore \mathrm{BD}=\mathrm{CE} \text {. } \\ \end{aligned}

(2) $\angle \mathrm{AEB}=90^{\circ}, \mathrm{AE}=\mathrm{BE}+2 \mathrm{CM}$, ,

\begin{aligned} & \triangle \mathrm{ACD} \cong \triangle \mathrm{BCE}, \\ & \therefore \mathrm{AD}=\mathrm{BE}, \angle \mathrm{ADC}=\angle \mathrm{BEC}, \\ & \because \triangle \mathrm{CDE} \text { 是等腰直角三角形, } \\ & \therefore \angle \mathrm{CDE}=\angle \mathrm{CED}=45^{\circ}, \\ & \therefore \angle \mathrm{ADC}=180^{\circ}-\angle \mathrm{CDE}=135^{\circ}, \\ & \therefore \angle \mathrm{BEC}=\angle \mathrm{ADC}=135^{\circ}, \\ & \therefore \angle \mathrm{AEB}=\angle \mathrm{BEC}-\angle \mathrm{CED}=135^{\circ}-45^{\circ}=90^{\circ} . \\ & \because \mathrm{CD}=\mathrm{CE}, \mathrm{CM} \perp \mathrm{DE}, \end{aligned}

\begin{aligned} & \therefore \mathrm{DM}=\mathrm{ME} . \quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ & \because \angle \mathrm{DCE}=90^{\circ}, \\ & \therefore \mathrm{DM}=\mathrm{ME}=\mathrm{CM}, \quad \therefore \mathrm{DE}=2 \mathrm{CM} . \\ & \therefore \mathrm{AE}=\mathrm{AD}+\mathrm{DE}=\mathrm{BE}+2 \mathrm{CM} . \end{aligned}