Quadrangle $A B C D$ is inscribed in a circle with radius 1 in such a way that one diagonal, $A C$, is a diameter of the circle, while the other diagonal, $B D$, is as long as $A B$. The diagonals intersect in $P$. It is known that the length of $P C$ is $\frac{2}{5}$. How long is the side $C D$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_31b1db04c15387bffd81g-4.jpg?height=271&width=320&top_left_y=207&top_left_x=594)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_17_31b1db04c15387bffd81g-4.jpg?height=252&width=323&top_left_y=228&top_left_x=912)

Figure 2

![](https://cdn.mathpix.com/cropped/2024_04_17_31b1db04c15387bffd81g-4.jpg?height=271&width=234&top_left_y=224&top_left_x=1299)

Figure 3
Let $\angle A C D=2 \alpha$ (see Figure 1). Then $\angle C A D=\frac{\pi}{2}-2 \alpha, \angle A B D=2 \alpha, \angle A D B=\frac{\pi}{2}-\alpha$ and $\angle C D B=\alpha$. The sine theorem applied to triangles $D C P$ and $D A P$ yields

$$
\frac{|D P|}{\sin 2 \alpha}=\frac{2}{5 \sin \alpha}
$$

and

$$
\frac{|D P|}{\sin \left(\frac{\pi}{2}-2 \alpha\right)}=\frac{8}{5 \sin \left(\frac{\pi}{2}-\alpha\right)}
$$

Combining these equalities we have

$$
\frac{2 \sin 2 \alpha}{5 \sin \alpha}=\frac{8 \cos 2 \alpha}{5 \cos \alpha}
$$

which gives $4 \sin \alpha \cos ^{2} \alpha=8 \cos 2 \alpha \sin \alpha$ and $\cos 2 \alpha+1=4 \cos 2 \alpha$. So we get $\cos 2 \alpha=\frac{1}{3}$ and $|C D|=$ $2 \cos 2 \alpha=\frac{2}{3}$.