1. The area of a right-angled triangle is $45 \mathrm{~cm}^{2}$. Find the area of the triangle whose vertices are the orthogonal projections of the centroid onto the sides of the triangle.
Solution. Let $D, E$ and $F$ be the orthogonal projections of the centroid $T$ of triangle $ABC$ onto the sides $AC, BC$ and $AB$, respectively. We will prove that $P_{\triangle DEF}=\frac{2}{3} P$, where $P$ is the area of triangle $ABC$. Clearly, quadrilateral $CDTE$ is a rectangle, so according to Thales' theorem

![](https://cdn.mathpix.com/cropped/2024_06_05_bd3f37f65d5d6e754f63g-04.jpg?height=408&width=661&top_left_y=534&top_left_x=818)
we get:

$$
\overline{CD}: \overline{CA}=\overline{A_1 T}: \overline{AA_1}=1: 3, \overline{CE}: \overline{CB}=\overline{B_1 T}: \overline{BB_1}=1: 3
$$

From this, it follows that $DE \parallel AB$ and $\overline{DE}=\frac{1}{3} \overline{AB}$. If $FP \perp DE$, i.e., $FP \perp AF$, then $FP \parallel CH$ and $\overline{FP}=\frac{2}{3} \overline{CH}$. For the area of triangle $\triangle DEF$, we get

$$
P_{DEF}=\frac{1}{2} \overline{DE} \cdot \overline{FP}=\frac{1}{2} \cdot \frac{1}{3} \overline{AB} \cdot \frac{2}{3} \overline{CH}=\frac{2}{9} P
$$

In our case, we get $P_{DEF}=\frac{2}{9} \cdot 45=10$.

Thus, the required area is $10 \mathrm{~cm}^{2}$.