Croatian Mathematical Olympiad

Let $M$ and $N$ be the midpoints of the segments $\overline{CX}$ and $\overline{DX}$, respectively. Also, denote $\alpha =\angle BAC=\angle BDC$ and $\beta =\angle CAD=\angle CBD$.



Since the triangles $AXN$ and $ADN$ have equal areas, we have $|AX|\cdot |AN|\sin \alpha =|AD|\cdot |AN|\sin \beta$, from which $$\frac{|AD|}{|AX|}=\frac{\sin \alpha }{\sin \beta }=\frac{|BC|}{|CD|},\text{i.e.}\frac{|AX|}{|CD|}=\frac{|AD|}{|BC|}$$

follows. Analogously, we get $\frac{|BX|}{|CD|}=\frac{|BC|}{|AD|}$. By the inequality of arithmetic and geometric means, it follows that $$\frac{|AB|}{|CD|}=\frac{|AX|}{|CD|}+\frac{|BX|}{|CD|}=\frac{|AD|}{|BC|}+\frac{|BC|}{|AD|}\ge 2.$$ The value 2 is attained for an isosceles trapezium $ABCD$ such that $|AB|=2|CD|$ and $X$ being the midpoint of the base $\overline{AB}$.