Example 37. We will find the length of the diagonal of a square, two vertices of which lie on a circle of radius $R$, and the other two - on a tangent to this circle.

Given (Fig. 41): $\omega(0,|O A|)$ - circle, $|O A|=R$, $A B C D$ - square, $A \in \omega, D \in \omega,(B C)$ - tangent to $\omega,[A C]-$ diagonal of the square.

To find: $|A C|$.
S o l u t i o n. It is clear that this is a problem of type $F$. We will use the algebraic method to solve it. Let the line $B C$ touch the circle $\omega$ at point $M$, and the line $C D$ intersect it at point $F$. Draw $[OM]$, $[O F]$, and $[O K]$, where $[O K] \perp[C D]$. Note that

$$
|C F| \cdot|C D|=|C M|^{2}
$$

Let $|C D|=2 x$. Since $O M C K-$ is a rectangle, then $|M C|=|O K|=x$, and $|C K|=|O M|=R$.

From the right triangle OFK, we find:

$$
|F K|=\sqrt{R^{2}-x^{2}}
$$

Since $|C F|=|C K|-|F K|$, then

$$
|C F|=R-\sqrt{R^{2}-x^{2}} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-063.jpg?height=528&width=462&top_left_y=244&top_left_x=360)

Fig. 42

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-063.jpg?height=531&width=460&top_left_y=862&top_left_x=364)

Fig. 43

Substituting the values of $|C F|,|C D|$, and $|C K|$ into equation (37.1), we get the equation

$$
\left(R-\sqrt{R^{2}-x^{2}}\right) 2 x=x^{2},
$$

solving which we find $x_{1}=0$ and $x_{2}=\frac{4}{5} R$.

Investigation. By the problem's meaning, $0<x<R$, so $|C D|=\frac{8}{5} R$. Therefore, $|A C|=\frac{8 \sqrt{ }{ }^{2}}{5} R$.

Example 38. A quadrilateral $A B C D$ is inscribed in a circle of radius $R$, and its diagonals intersect at point $E$ such that $|A E|:|E C|=2: 3$. Find the side $C D$ if it is known that triangle $A B C$ is equilateral.

Solution. Since triangle $A B C$ is equilateral (Fig. 42), then $|A C|=$ $=R \sqrt{3}$.

Since angles $B D C$ and $B A C$ are inscribed angles subtending the same arc of the circle $\omega$, then $\widehat{B D C}=\widehat{B A C}=60^{\circ}$. Similarly, $\widehat{B D A}=$ $=\widehat{B C A}=60^{\circ}$. Therefore, in triangle $A D C$, $\widehat{A D C}=120^{\circ}$, and $[D E)$ is the angle bisector of $\widehat{A D C}$, so $|A D|:|C D|=|A E|:|E C|$, i.e., $|A D|:|C D|=2: 3$. Let $|C D|=x$, then $|A D|=\frac{2}{3} x$.

By the cosine rule for triangle $A C D$, we have:

$$
|A C|^{2}=|A D|^{2}+|C D|^{2}-2|A D| \cdot|C D| \cdot \cos \widehat{A D C} .
$$

Substituting the values of $|A C|,|A D|,|C D|$, and $\cos \widehat{A D C}=\cos 120^{\circ}=-\frac{1}{2}$ into this equation, we get the equation

$$
3 R^{2}=\frac{4}{9} x^{2}+x^{2}+\frac{2}{3} x^{2}
$$

solving which we find:

$$
x_{1,2}= \pm \frac{3 R \sqrt{57}}{19}
$$

Investigation. By the problem's meaning, $0<x<R \sqrt{3}$. 

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-063.jpg?height=91&width=394&top_left_y=2236&top_left_x=357)

Therefore, $|C D|=\frac{3 R \sqrt{57}}{19}$.