2003 Vietnamese Mathematical Olympiad

1/ Let $K$ be the point of intersection of $MD$ and $NE$. It is clear that the circle with diameter $BK$ circumscribes about $△BED$.

We have: $AH\parallel MK$ and $CH\parallel NK$.

It implies: $\angle HAC=\angle KMN$ and $\angle ACH=\angle MNK$.

From these equalities and from $AC=MN$, we see that $△AHC\cong △MKN$. Thus $d(K,AC)=d(H,AC)$. But $K$ and $H$ lie on the same side with respect to the line $AC$, it shows that $KH\parallel AC$. Therefore $KH\perp BH$ and $H$ lies on the circle with diameter $BK$ circumscribing about $△BED$.

2/ From the proof of 1/, it is clear that $O^{\prime }$ is the midpoint of segment $BK$. Let $I$ be the midpoint of segment $AN$. It is easy to see that the orthogonal projections of $I$ on the line $BA$ and on the line $BC$ are respectively the midpoints of $EA$ and $DC$. Therefore, the orthogonal projections of vector $O^{\prime }M$ on the line $BA$ and on the line $BC$ are respectively $(1/2)\overrightarrow{BA}$ and $(1/2)\overrightarrow{BC}$. It shows that $\overrightarrow{OM}=\overrightarrow{BO}$. Thus, quadrilateral $B{O}^{\prime }MO$ is a parallelogram and the points $B$ and $M$ are symmetric with respect to the midpoint of segment $O{O}^{\prime }$.