Let $D$ be a point on the side $[AB]$ of the isosceles triangle $ABC$ such that $|AB|=|AC|$. The parallel line to $BC$ passing through $D$ intersects $AC$ at $E$. If $m(\widehat A) = 20^\circ$, $|DE|=1$, $|BC|=a$, and $|BE|=a+1$, then which of the followings is equal to $|AB|$?

$ 
\textbf{(A)}\ 2a
\qquad\textbf{(B)}\ a^2-a
\qquad\textbf{(C)}\ a^2+1
\qquad\textbf{(D)}\ (a+1)^2
\qquad\textbf{(E)}\ a^2+a
$
1. Given that $ABC$ is an isosceles triangle with $|AB| = |AC|$ and $m(\widehat{A}) = 20^\circ$. Let $D$ be a point on $[AB]$ such that the line parallel to $BC$ passing through $D$ intersects $AC$ at $E$. We are given $|DE| = 1$, $|BC| = a$, and $|BE| = a + 1$.

2. Since $DE \parallel BC$, triangles $ADE$ and $ABC$ are similar by the Basic Proportionality Theorem (also known as Thales' theorem). This gives us the ratio:
   \[
   \frac{DE}{BC} = \frac{AD}{AB} = \frac{AE}{AC}
   \]
   Given $DE = 1$ and $BC = a$, we have:
   \[
   \frac{1}{a} = \frac{AE}{AC}
   \]

3. Let $BE$ and $CD$ meet at $F$. Since $DE \parallel BC$, $BF = CF = a$ because $F$ is the midpoint of $BC$ in the similar triangles.

4. Since $\angle A = 20^\circ$ and $ABC$ is isosceles with $|AB| = |AC|$, the base angles $\angle ABC$ and $\angle ACB$ are each:
   \[
   \angle ABC = \angle ACB = \frac{180^\circ - 20^\circ}{2} = 80^\circ
   \]

5. In triangle $ABE$, since $DE \parallel BC$, $\angle ABE = \angle ABC = 80^\circ$. Therefore, $\angle AEB = 180^\circ - \angle ABE - \angle BAE = 180^\circ - 80^\circ - 20^\circ = 80^\circ$.

6. Since $AE = BE = a + 1$ and $DE \parallel BC$, we use the similarity ratio:
   \[
   \frac{AE}{AC} = \frac{1}{a}
   \]
   Substituting $AE = a + 1$ and solving for $AC$:
   \[
   \frac{a + 1}{AC} = \frac{1}{a} \implies AC = a(a + 1) = a^2 + a
   \]

7. Since $|AB| = |AC|$, we have:
   \[
   |AB| = a^2 + a
   \]

The final answer is $\boxed{a^2 + a}$.