In the right-angled triangle $ABC$ at $C$, let $AB=c$, $AC=b$, and $CB=a$. The altitude from $C$ intersects the hypotenuse at point $D$. Let the segments $BD=x$ and $DA=y$.

Since the leg is the geometric mean of the hypotenuse and the segment adjacent to the leg, we can write:

$$
a^{2}=c x, \quad b^{2}=c y
$$

from which

$$
\begin{gathered}
a^{2}: b^{2}=x: y \\
a^{2}+b^{2}: a^{2}-b^{2}=x+y: x-y \\
c^{2}:(a+b) d=c: e
\end{gathered}
$$

from which

$$
(a+b) d=c e
$$

$$
\frac{a+b}{c}=\frac{e}{d}
$$

But

$$
\frac{a+b}{c}=\sin \alpha+\cos \alpha=\frac{\cos \left(\alpha-45^{\circ}\right)}{\sin 45^{\circ}}
$$

From (2) and (3) we get:

$$
\cos \left(\alpha-45^{\circ}\right)=\frac{e \sin 45^{\circ}}{d}
$$

from which $\alpha-45^{\circ}=32^{\circ} 19^{\prime}$, so $\alpha=77^{\circ} 19^{\prime}$.

Once we know the angles, we can easily calculate the sides. Since

$$
a=b \tan \alpha \text { and } a-b=d
$$

we get:

$$
d=b(\tan \alpha-1)
$$

from which

$$
b=\frac{d}{\tan \alpha-1}
$$

But

$$
\tan \alpha-1=\frac{\sin \alpha-\cos \alpha}{\cos \alpha}=\frac{\sin \left(\alpha-45^{\circ}\right)}{\cos \alpha \cos 45^{\circ}}
$$

so

$$
b=\frac{d \cos \alpha \cdot \cos 45^{\circ}}{\sin \left(\alpha-45^{\circ}\right)}
$$

from which

$$
b=9
$$

And since $a-b=d$, we get:

$$
\begin{gathered}
a=b+d=40 \\
c=\sqrt{a^{2}+b^{2}}=41
\end{gathered}
$$

Thus, the sides of the triangle are: $a=40 \mathrm{~cm}, b=9 \mathrm{~cm}, c=41 \mathrm{~cm}$.

(Aladár Visnya, realist VIII.o.t., Pécs.)

The problem was also solved by: Ervin Hofbauer and Nándor Kántor, Budapest; István Szabó, Debrecen; Bernát Friedmann, S.A.-Ujhely; Béla Grünhut, Pécs.
In a right-angled triangle, the height dropped to the hypotenuse divides the hypotenuse into two segments whose difference is $e$; the difference between the two legs is $d$. What are the sides and angles of the triangle?

For example:

$$
e=37.0488 \mathrm{~cm}, \quad d=31 \mathrm{~cm}
$$