Let's modify the right-angled triangle $ABC$ on the hypotenuse $AB$ to find the point $X$ such that $XA: XB = BC^2: AC^2$.
Let $X A=x$, thus with the usual notation $X B=c-x$, so from the requirement

$$
x:(c-x)=a^{2}: b^{2}, \quad b^{2} x=a^{2} c-a^{2} x
$$

considering that the triangle is right-angled,

$$
x=\frac{a^{2} c}{a^{2}+b^{2}}=\frac{a^{2} c}{c^{2}}=\frac{a^{2}}{c}, \quad \text { or } \quad c x=a^{2}
$$

Accordingly, the leg $a$ is the geometric mean between the sought segment $x$ and the hypotenuse. It is known, however, that the leg $a$ is the geometric mean between the projection $a^{\prime}$ of the hypotenuse and the hypotenuse itself: $a^{2}=a^{\prime} c$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4bea76fa53bf3edaa93bg-1.jpg?height=293&width=466&top_left_y=585&top_left_x=816)

From these, $x=a^{\prime}$, that is, if the projection of $C$ on $A B$ is denoted by $C^{\prime}$, then $X A=B C^{\prime}$. Thus, $X$ is obtained by taking the reflection of $C^{\prime}$ over the midpoint $C_{0}$ of the hypotenuse $A B$.

Maria Tasnady (Budapest, Fazekas M. High School)

Remark. Those who used the quoted theorem in reverse wrote the requirement as $X A: X B=\left(B C^{\prime} \cdot B A\right):\left(A C^{\prime} \cdot A B\right)=B C^{\prime}: A C^{\prime}$ and derived from this, relying on $X A+X B=B C^{\prime}+A C^{\prime}$, the result $X A=B C^{\prime}, X B=A C^{\prime}$.