If in a right-angled triangle we draw the midlines $K_{1}$ and $K_{2}$ corresponding to the legs $a$ and $b$, then

$$
16 K_{2}^{2}-4 K_{1}^{2}=15 a^{2}
$$
$$
\begin{aligned}
& K_{1}^{2}=\frac{a^{2}}{4}+b^{2} \\
& K_{2}^{2}=a^{2}+\frac{b^{2}}{4}
\end{aligned}
$$

so

$$
16 K_{2}^{2}-4 K_{1}^{2}=16 a^{2}+4 b^{2}-\left(a^{2}+4 b^{2}\right)=15 a^{2} .
$$

(Basch Rudolf, Budapest.)