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An $a\times a$ square intersects two parallel lines located at distance $a$ from each other. The points of intersection are cross joined. Prove that the angle so obtained is $45^{\circ}.$

The problem is credited to V. V. Proizvolov (Kvant 1996 (2)) and may serve an example where using dynamical software to sketch a diagram proves to be a distraction. I put together a GeoGebra applet and looked at the possible properties of the configuration with several elements added. Meanwhile, Hubert Shutrick pointed out its obvious, salient feature. The solution is a one liner.

Solution

The angle formed by a transversal with two parallel lines is uniquely determined by its length between the lines - and vice versa.

The rest is easy:

$\gamma+\delta=90^{\circ}$

$2\alpha + 2\beta +\gamma+\delta=360^{\circ}$

$\alpha + \beta =135^{\circ}$

$\epsilon =45^{\circ}$