As points $A, B, C, D$ move in the plane, $A D=B C=2$ and $A C=B D=4$ always hold, furthermore, segments $A C$ and $B D$ intersect. How does the distance $C D$ depend on the distance $A B$?
I. solution. The triangles $A B C$ and $A B D$ are congruent because their side $A B$ is common, and their other two sides are equal. Therefore, $\angle A C B = \angle A D B = \alpha$. Since $A C$ and $B D$ intersect, $C$ and $D$ are on the same side of the line $A B$. Because of the equality of the angles, $C$ and $D$ lie on the same circle of the two circles of view corresponding to the angle $\alpha$ of the segment $A B$, so $A B C D$ is a cyclic quadrilateral.

![](https://cdn.mathpix.com/cropped/2024_05_02_5a6f93427c5882e308f1g-1.jpg?height=241&width=596&top_left_y=351&top_left_x=754)

According to Ptolemy's theorem (see, for example, Problem 1259 in the Collection of Geometric Problems I.), the sum of the products of the opposite sides of any cyclic quadrilateral is equal to the product of the diagonals. In our case, therefore,

$$
A B \cdot C D + A D \cdot B C = A C \cdot B D
$$

that is,

$$
C D = \frac{A C \cdot B D - A D \cdot B C}{A B} = \frac{4 \cdot 4 - 2 \cdot 2}{A B} = \frac{12}{A B}
$$

By the triangle inequality, $A B > A C - B C = 4 - 2 = 2$ and $A B < A C + B C = 4 + 2 = 6$. When these conditions are met, $2 < C D < 6$ also holds, and in such a case, the quadrilateral $A B C D$ exists.

II. solution. The cyclic quadrilateral $A B C D$ is an isosceles trapezoid because two of its opposite sides, $A D$ and $B C$, are equal. Therefore, $\angle B A D + \angle C D A = 180^{\circ}$, or $\cos \angle B A D + \cos \angle C D A = 0$. Applying the cosine rule to triangles $A B D$ and $A C D$,

$$
\cos \angle B A D = \frac{A B^2 + A D^2 - B D^2}{2 \cdot A B \cdot A D} \quad \text{and} \quad \cos \angle C D A = \frac{A D^2 + C D^2 - A C^2}{2 \cdot A D \cdot C D},
$$

so

$$
\frac{A B^2 + 2^2 - 4^2}{4 \cdot A B} + \frac{2^2 + C D^2 - 4^2}{4 \cdot C D} = 0
$$

From this, after multiplying through and rearranging, we get

$$
(A B + C D)(A B \cdot C D - 12) = 0
$$

Since $A B + C D \neq 0$, it follows that $(A B \cdot C D - 12) = 0$, or $C D = 12 / A B$.