Solution. Let $M$ be the point of intersection of the tangents. By the condition, $M$ lies on the line $B D$. Since $B C < C D$, point $B$ is between points $M$ and $D$. Then, since $\angle M C B = \angle C D B = \frac{\smile B C}{2}$, triangles $M C B$ and $M D C$ are similar, so $\frac{C D}{B C} = \frac{M D}{M C}$. Similarly, from the similarity of triangles $M A B$ and $M D A$, we get $\frac{A D}{A B} = \frac{M D}{A M}$. Moreover, the tangent segments to the circle $M C$ and $M A$ are equal, so $\frac{C D}{B C} = \frac{A D}{A B}$, which means $A D = \frac{A B \cdot C D}{B C} = \frac{2 \cdot 5}{4} = \frac{5}{2}$.

Answer to variant 2: $\frac{9}{4}$. Answer to variant 3: $\frac{5}{4}$. Answer to variant 4: $\frac{7}{2}$.
4. Quadrilateral $ABCD$ is inscribed in a circle. The tangents to this circle, drawn at points $A$ and $C$, intersect on the line $BD$. Find the side $AD$, if $AB=2$ and $BC: CD=4: 5$.

Answer: $\frac{5}{2}$.