## Task 3.

On the side $\overline{A B}$ of the cyclic quadrilateral $A B C D$, there is a point $X$ with the property that the diagonal $\overline{B D}$ bisects the length of $\overline{C X}$, and the diagonal $\overline{A C}$ bisects the length of $\overline{D X}$.

What is the smallest possible ratio $|A B|:|C D|$ in such a quadrilateral?
## First Solution.

Let $M$ and $N$ be the midpoints of segments $\overline{C X}$ and $\overline{D X}$, and let $\alpha=\varangle B A C=\varangle B D C$ and $\beta=\varangle C A D=$ $\varangle C B D$.

![](https://cdn.mathpix.com/cropped/2024_05_30_ac9e556518bacc93315ag-08.jpg?height=702&width=671&top_left_y=397&top_left_x=698)

Then $P(A X N)=P(A D N)$, so $|X A| \cdot|A N| \sin \alpha=|A D| \cdot|A N| \sin \beta$, which implies

$$
\frac{|X A|}{|A D|}=\frac{\sin \alpha}{\sin \beta}=\frac{|D C|}{|B C|}
$$

Rewriting, we get $\frac{|X A|}{|C D|}=\frac{|A D|}{|B C|}$. Similarly, we obtain $\frac{|X B|}{|C D|}=\frac{|B C|}{|A D|}$.

Now it follows that

$$
\frac{|X A|}{|C D|}+\frac{|X B|}{|C D|}=\frac{|A D|}{|B C|}+\frac{|B C|}{|A D|} \geqslant 2
$$

Equality holds if $A B C D$ is an isosceles trapezoid where $|A B|=2|C D|$ and $X$ is the midpoint of segment $\overline{A B}$.