The tangent line drawn through vertex $C$ of the inscribed triangle $ABC$ intersects

the extension of side $AB$ beyond vertex $B$ at point $D$. It is known that the radius of the circle is $2, AC=\sqrt{12}$, and $L$

$\angle CDA + \angle ACB = 2 \angle BAC$. Find the secant $AD$.

#
From the Law of Sines, it follows that $\angle A B C=60^{\circ}$ or $\angle A B C=120^{\circ}$. By applying the theorem about the angle between a tangent and a chord, prove that $\angle B A C=45^{\circ}$. Then find the other angles of triangle $A D C$ and use the Law of Sines.

## Solution

Let $\angle B A C=\alpha, \angle A B C=\beta, \angle A C B=\gamma, \angle C D A=\varphi$. Let $R=\sqrt{12}=2 \sqrt{3}$ be the radius of the circumcircle of triangle $A B C$. Then

$$
\sin \beta=\sin \angle A B C=\frac{A C}{2 R}=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}
$$

Therefore, $\beta=60^{\circ}$ or $\beta=120^{\circ}$.

By the theorem about the angle between a tangent and a chord, $\angle B C D=\angle B A C=\alpha$. According to the problem, $\gamma=2 \alpha-\varphi$, and since by the theorem about the exterior angle of a triangle, $\beta=\varphi+\alpha$, substituting the expressions for $\gamma$ and $\beta$ into the equation $\beta+\gamma+\alpha=180^{\circ}$, we get the equation in terms of $\alpha: 4 \alpha=180^{\circ}$, from which we find that $\alpha=45^{\circ}$.

If $\beta=60^{\circ}$, then

$$
\varphi=\beta-\alpha=60^{\circ}-45^{\circ}=15^{\circ}, \gamma=75^{\circ}, \angle A C D=\alpha+\gamma=120^{\circ}
$$

By the Law of Sines,

$$
\frac{A D}{\sin \angle A C D}=\frac{A C}{\sin \angle A D C}, \text { or } \frac{A D}{\sin 120^{\circ}}=\frac{2 \sqrt{3}}{\sin 15^{\circ}}
$$

Therefore,

$$
A D=\frac{2 \sqrt{3}}{\sin 15^{\circ}} \cdot \sin 120^{\circ}=\frac{3}{\sin 15^{\circ}}
$$

If $\beta=120^{\circ}$, then similarly we find that $A D=\frac{3}{\sin 75^{\circ}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-48.jpg?height=294&width=698&top_left_y=1505&top_left_x=680)

Answer

$A D=\frac{3}{\sin 15^{\circ}}$ or $A D=\frac{3}{\sin 75^{\circ}}$.