Let's construct a right-angled triangle, given the hypotenuse $c$ and knowing that the median to the hypotenuse is the geometric mean between the two legs.
I. solution: It does not harm the generality if we assume once and for all that $a > b$. According to the Pythagorean theorem,

$$
a^{2}+b^{2}=c^{2}
$$

According to the condition (since the median is $\frac{c}{2}$),

$$
2 a b=\frac{c^{2}}{2}
$$

Adding and subtracting (1) and (2) we get

$$
\begin{aligned}
& (a+b)^{2}=\frac{3 c^{2}}{2} \\
& (a-b)^{2}=\frac{c^{2}}{2}
\end{aligned}
$$

from which

$$
\begin{aligned}
& a+b=\frac{c \sqrt{3}}{\sqrt{2}}=\frac{c \sqrt{6}}{2} \\
& a-b=\frac{c \sqrt{2}}{2}
\end{aligned}
$$

Based on these, the construction can be done in several ways.

$\alpha)$ Adding (3) and (4) we get that $a=\frac{c \sqrt{6}+c \sqrt{2}}{4}$, which is a segment that can be easily constructed ( $c \sqrt{6}$ is the geometric mean between $2c$ and $3c$, and $c \sqrt{2}$ is between $c$ and $2c$). With $a$ and $c$ known, the construction of the desired triangle is trivial.

$\beta)$ According to (3), $a+b$, as the geometric mean between $c$ and $\frac{3}{2}c$, can be constructed. With the segments $a+b$ and $c$ known, the construction of the triangle is familiar from the textbook (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_17d03ff38950c9d9d847g-1.jpg?height=534&width=483&top_left_y=1408&top_left_x=802)

Fig. 1

ү) According to (4), $a-b$ is the diagonal of a square whose side is $\frac{c}{2}$. With $a-b$ and $c$ known, the right triangle can be constructed in a known way (Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_17d03ff38950c9d9d847g-1.jpg?height=461&width=488&top_left_y=2168&top_left_x=797)

Fig. 2

In all three cases, it can be easily shown that the constructed triangle indeed satisfies our condition. For example, in the last case, according to the Pythagorean theorem,

$$
(a-b)^{2}=a^{2}+b^{2}-2 a b=2 \cdot \frac{c^{2}}{4}=\frac{c^{2}}{2}
$$

from which

$$
2 a b=a^{2}+b^{2}-\frac{c^{2}}{2}=c