Quadrilateral $ABCD$ is inscribed in a circle, $I$ is the center of the inscribed circle of triangle $ABD$. Find the minimum value of $BD$, if $AI=BC=CD=2$.
From the condition, it follows that the arcs $BC$ and $CD$ are equal, which means that the angle bisector $AI$ of angle $BAD$ intersects the circle at point $C$ (see figures). By the trident lemma,

$CI = CB = CD = 2$. We can reason in different ways from here.

First method. Let $K$ and $N$ be the feet of the perpendiculars dropped from points $C$ and $I$ to $BD$ and $AD$ respectively (left figure). Then, from the equality of the right triangles $BKC$ and $ANI$, it follows that $CK = IN$. The perpendicular $IP$ to the diagonal $BD$ is also equal to $IN$ (these are the radii of the inscribed circle of triangle $ABD$), so $BD$ intersects the segment $CI$ at its midpoint $L$. Thus, $CL = IL = 1$.

By the theorem on the product of segments of chords, $BL \cdot DL = AL \cdot CL$. Let $BL = x$ and $DL = y$, then $xy = 3$. By the Cauchy inequality, the minimum value of the sum $x + y$ is achieved when $x = y = \sqrt{3}$.

Therefore, the value $BD = 2\sqrt{3}$ is the smallest possible.
![](https://cdn.mathpix.com/cropped/2024_05_06_5ff42d6d57f9dccbb1f5g-02.jpg?height=788&width=1156&top_left_y=2038&top_left_x=473)

Second method. By Ptolemy's theorem (see problem 52468), $AD \cdot BC + AB \cdot CD = AC \cdot BD$, that is, $2AD + 2AB = 4BD$. Therefore, $BD = \frac{1}{2}(AB + AD)$ (a triangle $ABD$ with this property is called a difference triangle).

Without loss of generality, we can assume that $AB \geq AD$. Draw a circle with center $C$ and radius 2, which intersects side $AB$ at point $E$ (right figure). The angle bisector $AC$ of angle $BAD$ is its axis of symmetry and the axis of symmetry of the angle, so points $E$ and $D$ are symmetric with respect to $AC$. Therefore, $AE = AD$.

By the theorem on the product of segments of secants, $AE \cdot AB = AI \cdot AF$ (IF is the diameter of the constructed circle).

Therefore, $AD \cdot AB = 2 \cdot 6 = 12$.

By the Cauchy inequality, $BD = \frac{1}{2}(AB + AD) \geq \sqrt{AB \cdot AD} = 2\sqrt{3}$. Equality is achieved if $AB = AD$.

## Answer

$2\sqrt{3}$.