## G3

Let $C D \perp A B(D \in A B), D M \perp A C(M \in A C)$ and $D N \perp B C(N \in B C)$ for an acute triangle ABC with area $S$. If $H_{1}$ and $H_{2}$ are the orthocentres of the triangles $M N C$ and MND respectively. Evaluate the area of the quadrilateral $\mathrm{AH}_{1} \mathrm{BH}_{2}$.


Solution1. Let $O, P, K, R$ and $T$ be the midpoints 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 $O K \| D N$

![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-28.jpg?height=612&width=789&top_left_y=818&top_left_x=995)
(from $\triangle C D N$ ) 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 $\Delta T R H_{1} \cong \triangle K P O$ (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 $\mathrm{CH}_{1} \| \mathrm{PO}$.

Analogously, $\overline{D H_{2}}=2 \overline{P O}$ and $D H_{2} \| P O$. From $\overline{C H_{1}}=2 \overline{P O}=\overline{D H_{2}}$ and

![](https://cdn.mathpix.com/cropped/2024_06_05_eb51ab1a35cd98c2706cg-28.jpg?height=75&width=1581&top_left_y=1895&top_left_x=222)
$H_{1} H_{2} \| C D$. Therefore the area of the quadrilateral $A H_{1} B H_{2}$ is $\frac{\overline{A B} \cdot \overline{H_{1} H_{2}}}{2}=\frac{\overline{A B} \cdot \overline{C D}}{2}=S$.
