Let $f$ be a straight line, and let's draw the lines $e$ and $g$ on either side of it, each at a unit distance from $f$. Let $E, F$, and $G$ be points on the respective lines such that the triangle they form has a right angle at $F$. What is the length of the altitude from $F$ in this triangle?
Extend the segment $F G$ so that its intersection point with line $e$ is $E^{\prime}$. Since the distance between lines $e$ and $f$ is equal to the distance between lines $f$ and $g$, and since $e, f$, and $g$ are parallel, it follows that $G F = F E^{\prime}$. Thus, the triangles $E F G$ and $E F E^{\prime}$ are mirror images of each other with respect to the line $E F$. In the triangle $E F E^{\prime}$, the height corresponding to the hypotenuse is exactly the distance between the lines $e$ and $f$, which is one unit. Therefore, the height from $F$ in the triangle $E F G$ is also one unit.

![](https://cdn.mathpix.com/cropped/2024_05_02_5f66399f285801c88525g-1.jpg?height=439&width=288&top_left_y=374&top_left_x=924)

Tóth Szilveszter (Tata, Eötvös J. Gimn., 11. o.t.)

Note. The example is based on the following idea: Take a strip of paper. Fold one side at an arbitrary angle, then fold the other side alongside it so that the edges of the paper strip fit together. The height of the resulting right triangle will be equal to the width of the paper strip.