Let $ABC$ be a triangle with $m(\widehat A) = 70^\circ$ and the incenter $I$. If $|BC|=|AC|+|AI|$, then what is $m(\widehat B)$?

$ 
\textbf{(A)}\ 35^\circ
\qquad\textbf{(B)}\ 36^\circ
\qquad\textbf{(C)}\ 42^\circ
\qquad\textbf{(D)}\ 45^\circ
\qquad\textbf{(E)}\ \text{None of above}
$
1. Given that $m(\widehat{A}) = 70^\circ$ and $|BC| = |AC| + |AI|$, we need to find $m(\widehat{B})$.
2. Extend $AC$ past $A$ to $I'$ such that $AI = AI'$. This construction helps us use symmetry in the problem.
3. Notice that $\angle I'CI = \angle ACI = \angle ICB$ because $I$ is the incenter and $CI$ is the angle bisector of $\angle ACB$.
4. Since $CI' = AC + AI' = AC + AI = CB$, we have that $\triangle I'CI \cong \triangle BCI$ by the SAS Congruence Postulate (Side-Angle-Side).
5. From the congruence, it follows that $\angle IBC = \angle II'C$.
6. Since $\triangle I'AI$ is isosceles with $AI = AI'$, we have $\angle II'C = \angle II'A = \frac{\angle IAC}{2}$.
7. Therefore, $\angle B = 2 \angle IBC = 2 \angle II'C = 2 \left(\frac{\angle IAC}{2}\right) = \angle IAC$.
8. Since $\angle IAC = \frac{\angle A}{2} = \frac{70^\circ}{2} = 35^\circ$, we conclude that $\angle B = 35^\circ$.

The final answer is $\boxed{35^\circ}$