Saudi Arabian Mathematical Competitions

Denote by $x$ the side length of the square and construct $B B^{\prime} \perp a$, $C C^{\prime} \perp d$, where $B^{\prime} \in a$, $C^{\prime} \in d$. Let $\alpha$ be the angle defined by lines $a$ and $A B$.

In triangle $A B^{\prime} B$ we have $x = A B \sin \alpha$, and in triangle $C C^{\prime} D$ we have $x = C D \cos \alpha$. It follows
$$\left(\frac{x}{A B}\right)^{2} + \left(\frac{x}{C D}\right)^{2} = \sin^{2} \alpha + \cos^{2} \alpha = 1,$$
hence
$$x = \frac{A B \cdot C D}{\sqrt{A B^{2} + C D^{2}}}$$
and the conclusion follows.

Remark. We have $\tan \alpha = \frac{C D}{A B}$ and this formula shows how we can construct the lines $a$, $b$, $c$, $d$ through points $A$, $B$, $C$, $D$ in order to get the square. Let $B B^{\prime\prime} \perp A B$ and $B B^{\prime\prime} = C D$. Then construct the line $a = A B^{\prime\prime}$ and $d \perp a$. Finally, construct $b \parallel a$ and $c \parallel d$.

(Saleh Saeed Al-Gamdi and Abrar Abdulmanem Al-Shaikh).

In order to prove that the side length of the square does not depend on the length of segment $B C$ it suffices to draw parallel lines to line $A D$. Then the square is fixed; we get all possible configurations for the segment $B C$. The first position is $A^{\prime} D^{\prime}$, where points $B$, $C$ coincide with one vertex of the square.