Lemma. Let $ABC$ be a triangle with $|AB|^2 + |AB| \cdot |AC| = |BC|^2$. Then $m(\widehat{A}) = 2m(\widehat{C})$.
Proof: Let $D$ be an intersection of interior angle bisector of $\widehat{A}$ with $BC$. Then $|BD| = |AB| \cdot k$, $|CD| = |AC| \cdot k$. $|AB| \cdot (|AB| + |BC|) \cdot k = |BC|^2 \cdot k \Rightarrow |AB| \cdot |BC| = |BC|^2 \cdot k \Rightarrow |AB| = |BC|$. Since $|BD| = |AB| \cdot k$, $\triangle ABD \sim \triangle CBA \Rightarrow m(\widehat{BCA}) = m(\widehat{BAD}) = \frac{1}{2}m(\widehat{BAC})$. Done.

By the lemma, $|AD|^2 + |AD| \cdot |AB| = |BD|^2 \Rightarrow m(\widehat{BAD}) = 2m(\widehat{B})$ and $|AD|^2 + |AD| \cdot |AC| = |CD|^2 \Rightarrow m(\widehat{CAD}) = 2m(\widehat{C})$. Let $m(\widehat{B}) = \beta$ and $m(\widehat{C}) = \alpha$. Then $m(\widehat{A}) = 120^\circ$.

$|DE|^2 = |CD|^2 + |CD| \cdot |CE| \Rightarrow m(\widehat{DCE}) = 2m(\widehat{DEC})$ and $m(\widehat{DCE}) + m(\widehat{DEC}) = m(\widehat{ADC}) = 3\beta \Rightarrow m(\widehat{DCE}) = 2\beta$ and $m(\widehat{DEC}) = \beta$. Thus, $A$, $B$, $E$, $C$ are concyclic.

Since $\alpha = 60^\circ - \beta$, the measure of the arc $\widehat{ACE}$ is equal to $240^\circ - 2\beta$. Let us take a point $F$ on the arc $\widehat{BEC}$ satisfying $m(\widehat{BAF}) = m(\widehat{CAF}) = 60^\circ$. The measure of the arc $\widehat{ABF}$ is equal to $240^\circ$. Therefore, $|AF| = |AE|$.

$m(\widehat{BCF}) = m(\widehat{BAF}) = 60^\circ$, $m(\widehat{CBF}) = m(\widehat{CAF}) = 60^\circ$, $m(\widehat{BFC}) = 180^\circ - m(\widehat{BAC}) = 60^\circ \Rightarrow \triangle BCF$ is equilateral.

Ptolemy's cyclic quadrilateral theorem applied to $ABFC$ yields: $|AB| \cdot |CF| + |AC| \cdot |BF| = |AF| \cdot |BC|$. Therefore, $|AB| + |AC| = |AF|$ (since $|BF| = |CF| = |BC|$). Thus, $|AB| + |AC| = |AE|$. Done.