Consider an acute triangle $ABC$ with area $S$. Let $CD \perp AB \quad(D \in AB)$, $DM \perp AC \quad(M \in AC)$ and $\quad DN \perp BC \quad(N \in BC)$. Denote by $H_{1}$ and $H_{2}$ the orthocentres of the triangles $MNC$ and $MND$ respectively. Find the area of the quadrilateral $\mathrm{AH}_{1} \mathrm{BH}_{2}$ in terms of $S$.
1. Let $O, P, K, R$ and $T$ be the mid-points of the segments $C D, M N$, $C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ and $T R \| M C$. Consequently, $\overline{P K}=\overline{T R}$ and $P K \| T R$. Also $O K \| D N \quad$ (from

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$\triangle C D N$ ) and since $D N \perp B C$ and $M H_{1} \perp B C$, it follows that $T H_{1} \| O K$. Since $O$ is the circumcenter of $\triangle C M N, O P \perp M N$. Thus, $C H_{1} \perp M N$ implies $O P \| C H_{1}$. We conclude $\triangle T R H_{1} \cong \triangle K P O \quad$ (they have parallel sides and $\overline{T R}=\overline{P K}$ ), hence $\overline{R H_{1}}=\overline{P O}$, i.e. $\overline{C H_{1}}=2 \overline{P O}$ and $C H_{1} \| P O$.

Analogously, $\overline{D H_{2}}=2 \overline{P O} \quad$ and $\quad D H_{2} \| P O$. From $\quad \overline{C H_{1}}=2 \overline{P O}=\overline{D H_{2}} \quad$ and $C H_{1}\|P O\| D H_{2}$ the quadrilateral $C H_{1} H_{2} D$ is a parallelogram, thus $\overline{H_{1} H_{2}}=\overline{C D}$ and $H_{1} H_{2} \| C D$. Therefore the area of the quadrilateral $A H_{1} B H_{2}$ is $\frac{\overline{A B} \cdot \overline{H_{1} H_{2}}}{2}=\frac{\overline{A B} \cdot \overline{C D}}{2}=S$.

Solution 2. Since $M H_{1} \| D N$ and $N H_{1} \| D M, M D N H_{1}$ is a parallelogram. Similarly, $N H_{2} \| C M$ and $M H_{2} \| C N$ imply $M C N H_{2}$ is a parallelogram. Let $P$ be the midpoint of the segment $\overline{M N}$. Then $\sigma_{P}(D)=H_{1}$ and $\sigma_{P}(C)=H_{2}$, thus $C D \| H_{1} H_{2}$ and $\overline{C D}=\overline{H_{1} H_{2}}$. From $C D \perp A B$ we deduce $A_{A H_{1} B H_{2}}=\frac{1}{2} \overline{A B} \cdot \overline{C D}=S$.