Junior Balkan Mathematical Olympiad

Let $O$, $P$, $K$, $R$ and $T$ be the mid-points of the segments $CD$, $MN$, $CN$, $C{H}_1$ and $M{H}_1$, respectively. From $△MNC$ we have that $\overline{PK}=\frac{1}{2}\overline{MC}$ and $PK\parallel MC$.

Analogously, from $△M{H}_{1}C$ we have that $\overline{TR}=\frac{1}{2}\overline{MC}$ and $TR\parallel MC$. Consequently, $\overline{PK}=\overline{TR}$ and $PK\parallel TR$. Also $\overline{OK}\parallel \overline{DN}$ (from $△CDN$) and since $\overline{DN}\perp \overline{BC}$ and $\overline{M{H}_1}\perp \overline{BC}$, it follows that $\overline{T{H}_1}\parallel \overline{OK}$. Since $O$ is the circumcenter of $△CMN$, $\overline{OP}\perp \overline{MN}$. Thus, $C{H}_1\perp MN$ implies $\overline{OP}\parallel C{H}_1$. We conclude $△TR{H}_1\cong △KPO$ (they have parallel sides and $\overline{TR}=\overline{PK}$), hence $\overline{R{H}_1}=\overline{PO}$, i.e. $\overline{C{H}_1}=2\overline{PO}$ and $C{H}_1\parallel PO$.

Analogously, $\overline{D{H}_2}=2\overline{PO}$ and $D{H}_2\parallel PO$. From $\overline{C{H}_1}=2\overline{PO}=\overline{D{H}_2}$ and $C{H}_1\parallel PO\parallel D{H}_2$ the quadrilateral $C{H}_1{H}_{2}D$ is a parallelogram, thus $\overline{H_1{H}_2}=\overline{CD}$ and $H_1{H}_2\parallel CD$. Therefore the area of the quadrilateral $A{H}_{1}B{H}_2$ is $\frac{\overline{AB}\cdot \overline{H_1{H}_2}}{2}=\frac{\overline{AB}\cdot \overline{CD}}{2}=S$.

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Alternative solution.

Since $M{H}_1\parallel DN$ and $N{H}_1\parallel DM$, $MDN{H}_1$ is a parallelogram. Similarly, $N{H}_2\parallel CM$ and $M{H}_2\parallel CN$ imply $MCN{H}_2$ is a parallelogram. Let $P$ be the midpoint of the segment $MN$. Then ${\sigma }_P(D)=H_1$ and ${\sigma }_P(C)=H_2$, thus $CD\parallel H_1{H}_2$ and $\overline{CD}=\overline{H_1{H}_2}$. From $CD\perp AB$ we deduce $A_{A{H}_{1}B{H}_2}=\frac{1}{2}\overline{AB}\cdot \overline{CD}=S$.