Junior Balkan Mathematical Olympiad

Version 1. Let $I$ be the incenter of triangle $ABC$ and $R$ be the common point of the lines $BI$ and $MN$. Since $$m(\widehat{ANM})=90^{\circ }-\frac{1}{2}m(\widehat{MAN})\text{and}m(\widehat{BIC})=90^{\circ }+\frac{1}{2}m(\widehat{MAN})$$ 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 $M^{\prime }\in (AB)$ and $N^{\prime }\in (AC)$ be the unique points, such that $R\in M^{\prime }{N}^{\prime }$ and $(A{M}^{\prime })\equiv (A{N}^{\prime })$. (1'') Let $S$ be the intersection point of $CR$ and $AB$. Then $CR=RS$. (2'') Consider $K\in AC$ such that $SK\parallel M^{\prime }{N}^{\prime }$. Then $N^{\prime }$ is the midpoint of $(CK)$. (3'') We deduce $$A{N}^{\prime }=\frac{AK+KC}{2}=\frac{AS+AC}{2}=\frac{AB-BS+AC}{2}=\frac{AB+AC-BC}{2}=AN.(4^{\prime \prime })$$ We conclude that $N=N^{\prime }$, hence $M=M^{\prime }$, and $R,M,N$ are collinear. (5'')