Let's denote the orthocenter of triangle $ABC$ by $M$. How far is $M$ from the side $BC$, if the height drawn from $A$ is $15 \, \text{cm}$, and the side $BC$ is divided into segments of $10 \, \text{cm}$ and $6 \, \text{cm}$?
Solution. Let the foot of the altitude from vertex $A$ be $A_{1}$. Thus, $A A_{1}=15 \mathrm{~cm}, B A_{1}=10 \mathrm{~cm}, C A_{1}=6 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d3cd43b3a95777fd0a54g-1.jpg?height=464&width=496&top_left_y=226&top_left_x=804)

The triangle $B M A_{1}$ is similar to the triangle $A C A_{1}$, because both are right-angled, and one of their acute angles is equal, that is, $\angle M B A_{1}$ and $\angle C A A_{1}$ are complementary acute angles, so they are equal. Based on the similarity:

$$
\frac{M A_{1}}{B A_{1}}=\frac{C A_{1}}{A A_{1}}
$$

from which

$$
M A_{1}=\frac{B A_{1} \cdot C A_{1}}{A A_{1}}=\frac{10 \cdot 6}{15} \mathrm{~cm}=4 \mathrm{~cm}
$$

Based on the work of Mónika Horváth (Gyula, Erkel F. Gymnasium, 2nd grade)