Problem 1929

Solution:
Let $s = A M$. According to the conditions in the exercise we get:
$$s = A M = M B = \frac{1}{2} A B = A C = \frac{1}{2} C D = \frac{1}{3} A D$$

Using the power of the point $A$ with respect to the circle $E B D C$
$$A E = \frac{A C \cdot A D}{A B} = \frac{s \cdot 3s}{2s} = \frac{3s}{2}$$

and therefore
$$E B = A B - A E = 2s - \frac{3s}{2} = \frac{s}{2}$$

Since $F$ is an inner point of the triangle $A B D$, applying Ceva's theorem gives
$$1 = \frac{B X}{X D} \cdot \frac{D C}{C A} \cdot \frac{A E}{E B} = \frac{B X}{X D} \cdot \frac{2s}{s} \cdot \frac{3s/2}{s/2} = 6 \cdot \frac{B X}{X D} \implies \frac{B X}{X D} = \frac{1}{6}$$

Solution:
Using Thales' theorem over the circles $k$ and $E B C D$ gives
$$90^{\circ} = \angle A C B = \angle D C B = \angle D E B$$

Therefore, $D E$ and $B C$ are altitudes of the triangle $A B D$. Thus, $F$ is the orthocenter of $A B D$. It follows that $E B D C$, $A E X D$ and $A B X C$ are cyclic quadrilaterals (in particular $X \in k$).
By power of the points $A$, $B$, $D$ with respect to the circles $E B D C$, $A E X D$, $A B X C$ (individually) we get
$$\begin{aligned} & A E \cdot A B = A C \cdot A D \\ & B E \cdot B A = B X \cdot B D \\ & D X \cdot D B = D C \cdot D A \end{aligned}$$

Up to this point, we have not used any of the conditions about the lengths. Like in the first solution, we define $s = A M$ and get
$$s = A M = M B = \frac{1}{2} A B = A C = \frac{1}{2} C D = \frac{1}{3} A D$$

Finally
$$\frac{B X}{X D} = \frac{B X \cdot B D}{D X \cdot D B} = \frac{B E \cdot B A}{D C \cdot D A} = \frac{A B^{2} - A E \cdot A B}{D C \cdot D A} = \frac{A B^{2} - A C \cdot A D}{D C \cdot D A} = \frac{(2s)^{2} - s \cdot 3s}{2s \cdot 3s} = \frac{1}{6}$$