6. (8 points) Quadrilateral $ABCD$ is inscribed in a circle of radius $r$ such that diagonal $AC$ is the diameter of the circle. The diagonals of the quadrilateral intersect at point $P$. It is known that $BD = AB$ and $PC = \alpha < r$. Find the length of side $CD$.

#
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_873180233e487edb68a7g-06.jpg?height=762&width=757&top_left_y=344&top_left_x=661)

Let $O$ be the center of the circumscribed circle. Clearly, $A O = D O$ and $A B = D B$, so points $O$ and $B$ lie on the perpendicular bisector of segment $A D$. Since $A C$ is the diameter, $\angle A D C = 90^{\circ}$. Therefore, we get that $O B \| D C$. Consequently, triangle $O B P$ is similar to triangle $C D P$. Thus, we have

$$
\frac{C D}{O B} = \frac{C P}{O P} = \frac{C P}{C O - C P}
$$

From this, we can find the length of segment $C D$

$$
C D = \frac{O B \cdot C P}{C O - C P} = \frac{r \alpha}{r - \alpha}
$$

Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 1.68 | 2.08 | 2.52 | 0.375 | 0.96 | 1.9125 | 0.36 | 2.6125 | 0.5 | 1.05 | 2.99 |