[ Angles subtended by equal arcs and equal chords ]

$$
\text { [ } \quad \text { angle between a tangent and a chord }]
$$

An isosceles triangle \( A B C (A B = B C) \) is inscribed in a circle. The line \( C D \), perpendicular to \( A B \), intersects the circle at point \( P \). The tangent to the circle passing through point \( P \) intersects the line \( A B \) at point \( Q \). Find the segments \( P A \) and \( P Q \), if \( A C = 5 \) and \( \angle A B C = 2 \arccos \sqrt{\frac{5}{6}} \).
Let's assume that point $D$ lies on line $A B$. Let $B M$ be the height and median of triangle $A B C$. Denote $\angle A B M=\alpha$. Then

$$
\cos \alpha=\sqrt{\frac{5}{6}}, \sin \alpha=\sqrt{\frac{1}{6}}
$$

$$
\sin 2 \alpha=2 \sin \alpha \cos \alpha=2 \sqrt{\frac{5}{6}} \cdot \sqrt{\frac{1}{6}}=\frac{\sqrt{5}}{3}, \cos 2 \alpha=\sqrt{1-\frac{5}{9}}=\frac{2}{3}
$$

$$
\sin 3 \alpha=\sin (\alpha+2 \alpha)=\sin \alpha \cos 2 \alpha+\cos \alpha \sin 2 \alpha=\sqrt{\frac{1}{6}} \cdot \frac{2}{3}+\sqrt{\frac{5}{6}} \cdot \frac{\sqrt{5}}{3}=\frac{7 \sqrt{6}}{18}
$$

$$
\cos 3 \alpha=\sqrt{1-\frac{6 \cdot 49}{324}}=\frac{\sqrt{30}}{18}
$$

If $R$ is the radius of the circumcircle of triangle $A B C$, then

$$
R=\frac{A C}{2 \sin \angle A B C}=\frac{A C}{2 \sin 2 \pi 2}=\frac{5}{2 \cdot \frac{\sqrt{5}}{3}}=\frac{3 \sqrt{5}}{2}
$$

The inscribed angles $A B P$ and $A C P$ subtend the same arc, so

$$
\angle A B P=\angle A C P=\angle A B M=\alpha
$$

Thus,

$$
\angle P B C=\angle A B P+\angle A B C=\alpha+2 \alpha=3 \alpha
$$

From the theorem about the angle between a tangent and a chord,

$$
\angle C P Q=\angle P B C=3 \alpha
$$

Therefore,

$$
\begin{gathered}
P A=2 R \sin \angle A B P=2 R \sin \alpha=3 \sqrt{5} \cdot \frac{1}{\sqrt{6}}=\sqrt{\frac{15}{2}} \\
C P=2 R \sin \angle P B C=3 \sqrt{5} \cdot \frac{7 \sqrt{6}}{18}=\frac{7 \sqrt{30}}{6} \\
C D=A C \cos \angle A C D=A C \cos \alpha=5 \cdot \sqrt{\frac{5}{6}}=\frac{5 \sqrt{30}}{6}, \\
P D=C P-C D=\frac{7 \sqrt{30}}{6} - \frac{5 \sqrt{30}}{6}=\frac{\sqrt{30}}{3}
\end{gathered}
$$

Finally, from the right triangle $P D Q$, we find that

$$
P Q=\frac{P D}{\cos \angle C P Q}=\frac{P D}{\cos 3 \alpha}=\frac{\frac{\sqrt{30}}{3}}{\frac{\sqrt{30}}{18}}=6
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-13.jpg?height=1337&width=1146&top_left_y=1281&top_left_x=502)

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-14.jpg?height=174&width=1464&top_left_y=196&top_left_x=0)

A circumcircle is described around an isosceles triangle $A B C (A B=B C)$. The bisector of angle $B A C$ intersects the circle at point $D$. The tangent to the circle passing through point $D$ intersects line $A C$ at point $E$. Find the segments $C D$ and $D E$, if $A B=8, \angle B A C=2 \arcsin \frac{1}{\sqrt{3}}$.

## Answer

$2 \sqrt{6}, 5$.