In the right-angled triangle $ABC$, the hypotenuse $AB$ is divided in the ratio $m: n$, with the smaller segment being adjacent to the shorter leg. What ratio must exist between the sides of the right-angled triangle $ABC$ so that the lines connecting the division point to the midpoints of the legs form a right angle?
Let the foot of the perpendicular from $C$ to the hypotenuse $AB$ be $D$, such that

$$
A D: D B=m: n
$$

Let the midpoint of $AC=b$ be $F$, and the midpoint of $CB=a$ be $E$. Since $\angle ACB=90^{\circ}$, the line segment $FE$, which is the diameter of the circle drawn over it, passes through point $C$. Since this circle passes through the midpoints of two sides of the triangle and the feet of the altitudes from $C$ to $AC$ and $BC$, this circle is none other than the Feuerbach circle (Mathematical Practice Book II. 73.). It is known that this circle intersects the third side at its midpoint and at the foot of the altitude from $C$. Therefore, the sought point $D$ is the foot of the altitude from $C$. Consequently, we can write that

$$
A D: D B=m: n=\sqrt{b^{2}-C D^{2}}: \sqrt{a^{2}-C D^{2}}
$$

or, since

$$
C D=\frac{a b}{c}
$$

it follows that

$$
m: n=\sqrt{b^{2}-\frac{a^{2} b^{2}}{c^{2}}}: \sqrt{a^{2}-\frac{a^{2} b^{2}}{c^{2}}}=b \sqrt{c^{2}-a^{2}}: a \sqrt{c^{2}-b^{2}}
$$

and thus

$$
m: n=b^{2}: a^{2}
$$

(2) can also be written as

$$
m: n=c^{2}-a^{2}: a^{2}
$$

from which

$$
m+n: n=c^{2}: a^{2}
$$

or

$$
a: b: c=\sqrt{n}: \sqrt{m}: \sqrt{m+n}
$$

(Pál Sárközy, Pannonhalma.)