6. let $A, B, C$ and $D$ be four points lying in this order on a circle. Assume that there is a point $K$ on the line $A B$ such that $B D$ bisects the line $K C$ and $A C$ bisects the line $K D$. Determine the smallest possible value that $\left|\frac{A B}{C D}\right|$ can take.
Response: The minimum is 2.

Solution: We only use non-oriented (i.e., positive) distances.

We obtain a construction if $A B C D$ is an isosceles trapezoid with base $A B=2 C D$ and $K$ the midpoint of $A B$.

The condition on the midpoints implies that $[A D C]=[A K C]$ where the bracket notation denotes the area of the triangle. If we set $\alpha:=\angle C A D=\angle D B C$ and $\beta:=\angle B A C=\angle B D C$, we get

$$
\begin{aligned}
& {[A D C]=\frac{1}{2} \sin (\alpha) \cdot A C \cdot A D} \\
& {[A K C]=\frac{1}{2} \sin (\beta) \cdot A C \cdot A K}
\end{aligned}
$$

Thus, $A K / A D=\sin (\alpha) / \sin (\beta)$. Applying the Law of Sines to triangle $\triangle B C D$, we obtain

$$
\frac{\sin (\alpha)}{\sin (\beta)}=\frac{C D}{B C}
$$

and therefore $A K / C D=A D / B C$. By symmetry, $B C / C D=B C / A D$. Finally, using the AM-GM inequality:

$$
\frac{A B}{C D}=\frac{A K}{C D}+\frac{B K}{C D}=\frac{A D}{B C}+\frac{B C}{A D} \geq 2
$$