4. A circle is inscribed in an angle with vertex $A$, touching the sides of the angle at points $B$ and $C$. A line passing through $A$ intersects the circle at points $D$ and $E$. Chord $B X$ is parallel to line $D E$. In what ratio does the line $X C$ divide the chord $D E ?$

$(25$ points. $)$
Answer: $1: 1$.

First solution. (Fig. 2.) Let $O$ be the center of the circle, and $M$ be the intersection point of $C X$ and $D E$. We will prove that $O M \perp D E$, and thus, $D M = M E$.

First, the angle $\angle D M C$ is equal to half the sum of the arcs $C D$ and $E X$. Since the arcs between the parallel chords $B X$ and $D E$ are equal, $\overparen{D B} = \overparen{E X}$, therefore,

$$
\angle C M D = \frac{1}{2}(\overparen{C D} + \overparen{E X}) = \frac{1}{2}(\overparen{C D} + \overparen{D B}) = \frac{1}{2} \overparen{C B} = \frac{1}{2} \angle C O B = \angle C O A
$$

(The last equality follows, for example, from the congruence of right triangles $A O B$ and $A O C$.) From the equality of angles $\angle C O A$ and $\angle C M A$, it follows that points $A, C, M, O$ lie on the same circle. Since the radius $O C$ is perpendicular to the tangent $A C$, the diameter of this circle coincides with the segment $A O$. Therefore, $\angle A M O = 90^{\circ}$, which means $O M \perp D E$.

![](https://cdn.mathpix.com/cropped/2024_05_06_500bdf8839b2a3716788g-2.jpg?height=545&width=813&top_left_y=1427&top_left_x=252)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_500bdf8839b2a3716788g-2.jpg?height=526&width=789&top_left_y=1431&top_left_x=1145)

Fig. 3

Second solution. (Fig. 3.) Note that $\angle B C D = \angle E C X$, since the corresponding arcs are enclosed between the parallel chords. Additionally, from the equality of angles $\angle A B D$ and $\angle A E B$, it follows that triangles $A B D$ and $A E B$ are similar, and thus, $B D / B E = A D / A B$. Similarly, we obtain that $C D / C E = A D / A B$, which means $B D \cdot C E = C D \cdot B E = \frac{1}{2} B C \cdot D E$ (the last equality follows from Ptolemy's theorem.)

Now, let $C X$ intersect $D E$ at point $M$. Then triangles $C B D$ and $C M E$ are similar, so $B D \cdot C E = C B \cdot E M$. From this and the previous equality, we get that $E M = E D / 2$.

Criteria. It is proven that points $A, C, M, O$ lie on the same circle - 15 points.