denote $ \angle ABE = m, \angle CBE =n $ then $ \angle ABC =m+n= \angle ACB $, $ n= \angle CAE= \angle ABD $ $ \angle ACE = m $ Let we see the triangle $ ABD, \angle ADB =(180 - m - n) and \angle ABD = n $ then we now that $ \angle DAB = m = \angle ABE $ $ AD parallel to BE $

\[\begin{gathered} {\left. \begin{gathered} AB = AC \hfill \\ AD = CE \hfill \\ \end{gathered} \right\}_{_{_{}}}}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \left\{ \begin{gathered} arc.AB = arc.AC \hfill \\ arc.AD = arc.CE \hfill \\ \end{gathered} \right.\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} arc.BD = arc.AE\mathop {}\limits^{} \Rightarrow \hfill \\ \left. \begin{gathered} AD = CE \hfill \\ AB = AC \hfill \\ BD = AE \hfill \\ \end{gathered} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} tr.ABD = tr.ACE\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} A\hat CE = A\hat BE = D\hat AB = \varphi \mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} AD//BE \hfill \\ \end{gathered} \]

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Hmm here's an alternate solution. Note that by $AD=CE$ we have $\angle DBA=\angle EBC$, so $\angle DBE=\angle ABC=\angle ACB=\angle AEB$. Hence $EBDA$ is an isosceles trapezoid and $AD\parallel BE$ as desired.

Let $\angle BAC=x$ than $\angle ABC=\angle ACB=90-\frac{x} {2}$ and we have $AD=CE$ $\Rightarrow$ $\angle EBC=\angle ACD=y$ and $\angle EBA+\angle BAD=90-\frac{x} {2}+90+\frac{x} {2}$. That's why $BE$ parallel to $AD$. Done!

It suffices to show that $\angle DAB = \angle ABE$. Since $\angle ABC = \angle ACB$ we know that $\overarc{AB} = \overarc{AC}$ and since $\overline{AD} = \overline{EC}$ we know that $\overarc{AD} = \overarc{EC}$. Subtracting the two equations we have with arcs we see that $\overarc{DB} = \overarc{AE}$ and the result follows.

Angle chasing, we have $\angle EBC=\angle EDC=\angle EAC=\angle AED$. Let $P$ denote the intersection of $AC$ and $DE$. Now, $\triangle AEP \sim \triangle DCP$, so $\angle PCD=\angle CDP =\angle EBC$. This means that $\angle DCB=\angle ABE$. However, $\angle DCB=\angle DAB$, so $\angle ABE=\angle DAB$, implying that $DA$ and $BE$ are parallel.

Join $BD$ and $BE$. As $AD=CE$, we have that $\angle DBA=\angle CBE$. Let $\angle BAC= 2\theta$. Then, As $AB=AC, \angle ABC= 90^{\circ}-\theta=\angle ACB$. $ADBC$ is a cyclic quadrilateral $\implies \angle ACB+ \angle ADB= 180^{\circ}\implies \angle ADB= 90^{\circ}+\theta$. $\angle DBA=\angle CBE\implies \angle DBA+\angle ABE=\angle CBE+\angle ABE\implies \angle DBE=\angle ABC=90^{\circ}-\theta$. Then, $\angle ADB+\angle DBE=180^{\circ}$. Therefore, they are supplementary co-interior angles, implying that $AD\parallel BE$.

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Notice that $\overarc{BD} = \overarc{AE} \implies \angle BAD=\angle EBA \implies AD\parallel BE$