Version 1. Let $I$ be the incenter of triangle $ABC$ and $R$ be the common point of the lines $BI$ and $MN$. Since
$$m(\widehat{ANM}) = 90^\circ - \frac{1}{2} m(\widehat{MAN}) \quad \text{and} \quad m(\widehat{BIC}) = 90^\circ + \frac{1}{2} m(\widehat{MAN})$$
the quadrilateral $IRNC$ is cyclic. \qquad (1)
It follows that $m(\widehat{BRC}) = 90^\circ$ and therefore
$$m(\widehat{BCR}) = 90^\circ - m(\widehat{CBR}) = 90^\circ - (180^\circ - m(\widehat{BCD})) = \frac{1}{2} m(\widehat{BCD})$$
(2)

Version 2. If $R$ is the incenter of the trapezoid $ABCD$, then $B$, $I$ and $R$ are collinear, \qquad (1')
and $m(\widehat{BRC}) = 90^\circ$ \qquad (2')
The quadrilateral $IRNC$ is cyclic. \qquad (3')
Then $m(\widehat{MNC}) = 90^\circ + \frac{1}{2} \cdot m(\widehat{BAC})$ \qquad (4')
and $m(\widehat{RNC}) = m(\widehat{BIC}) = 90^\circ + \frac{1}{2} \cdot m(\widehat{BAC})$, \qquad (5')

Version 3. If $R$ is the incenter of the trapezoid $ABCD$, let $M' \in (AB)$ and $N' \in (AC)$ be the unique points, such that $R \in M'N'$ and $(AM') \equiv (AN')$. (1'')
Let $S$ be the intersection point of $CR$ and $AB$. Then $CR = RS$. (2'')
Consider $K \in AC$ such that $SK \parallel M'N'$. Then $N'$ is the midpoint of $(CK)$. (3'')
We deduce
$$AN' = \frac{AK + KC}{2} = \frac{AS + AC}{2} = \frac{AB - BS + AC}{2} = \frac{AB + AC - BC}{2} = AN. \quad (4'')$$
We conclude that $N = N'$, hence $M = M'$, and $R, M, N$ are collinear. (5'')