In a right-angled quadrilateral, whose sides are $A B=2 a, A B^{\prime}=2 a^{\prime}$, we take a point $C$, such that $B A C \angle=$ $B A^{\prime} C \angle=\alpha$. Calculate the distances of this point from the sides, as well as the angles $C B A$ and $C B^{\prime} A$.
From point $C$, perpendiculars are dropped to the sides $AB, BA', A'B'$, and $B'A$, yielding points $F, D, G$, and $E$. Let $CD = d$ and $CG = d'$. From triangle $CDA'$:

$$
d = d' \tan \alpha
$$

From triangle $CEA$:

$$
2a - d = (2a' - d') \cot \alpha
$$

Substituting (1) into (2):

$$
d' = \frac{2(a \tan \alpha - a')}{\tan^2 \alpha - 1}
$$

and from (1):

$$
d = \frac{2(a \tan \alpha - a' \tan \alpha)}{\tan^2 \alpha - 1}
$$

Furthermore:

$$
\begin{gathered}
CF = 2a' - d' = \frac{2(a' \tan^2 \alpha - a \tan \alpha)}{\tan^2 \alpha - 1} \\
CE = 2a - d = \frac{2(a' \tan \alpha - a)}{\tan^2 \alpha - 1}
\end{gathered}
$$

The angle $CBA$ is calculated from triangle $FCB$:

$$
\tan CBA = \frac{CF}{d} = \frac{a' \tan^2 \alpha - a \tan \alpha}{a \tan^2 \alpha - a' \tan \alpha} = \frac{a' \tan \alpha - a}{a \tan \alpha - a'}
$$

Similarly, the angle $CB'A$ is calculated from triangle $CB'E$:

$$
\tan CB'A = \frac{CE}{d'} = \frac{a' \tan \alpha - a}{a \tan \alpha - a'}
$$

We see that $\angle CBA = \angle CB'A$.

(Jenő Fekete.)

The problem was also solved by: F. Bojedain, M. Devecis, E. Freibauer, B. Friedmann, K. Goldziher, B. Kántor, N. Kántor, T. Kármán, Ö. Kornis, Gy. Krátky, H. Schiffer, Ö. Spitzer, I. Szabó, J. Weisz.