Saudi Arabia 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 $AB$.



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



Remark. We have $\tan \alpha =\frac{CD}{AB}$ 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 $B{B}^{\prime \prime }\perp AB$ and $B{B}^{\prime \prime }=CD$. Then construct the line $a=A{B}^{\prime \prime }$ and $d\perp a$. Finally, construct $b\parallel a$ and $c\parallel d$.

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Alternative solution.

(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 $BC$ it suffices to draw parallel lines to line $AD$. Then the square is fixed; we get all possible configurations for the segment $BC$. The first position is $A^{\prime }{D}^{\prime }$, where points $B$, $C$ coincide with one vertex of the square.