$\left[\begin{array}{l}{[\text { Equilateral (equilateral) triangle ] }} \\ {[\quad \text { Law of Cosines }}\end{array}\right]$

An equilateral triangle $ABC$ with side 3 is inscribed in a circle. Point $D$ lies on the circle, and the chord $AD$ equals $\sqrt{3}$. Find the chords $BD$ and $CD$.
Let point $D$ lie on the smaller arc $AB$ (Fig.1). Then the quadrilateral $ADBC$ is cyclic, therefore

$$
\angle ADB=180^{\circ}-\angle ACB=180^{\circ}-60^{\circ}=120^{\circ}
$$

Let $BD=x$. By the cosine rule,

$$
AB^2=AD^2+BD^2-2 AD \cdot BD \cos 120^{\circ}
$$

or $9=3+x^2+x\sqrt{3}$, from which we find that $BD=x=\sqrt{3}=AD$, and since $CA=CB$, the line $CD$ is the perpendicular bisector of the chord $AB$, hence $CD$ is the diameter of the circle. Therefore, $CD=2\sqrt{3}$.

If point $D$ lies on the smaller arc $AC$ (Fig.2), then similarly we find that $BD=2\sqrt{3}$ and $CD=\sqrt{3}$.

## Answer

$\sqrt{3}, 2\sqrt{3}$ or $2\sqrt{3}, \sqrt{3}$.