## Task A-1.2.

Given is the length of $\overline{A D}$ as 3. Let $B$ and $C(C \neq A)$ be points on the circle with diameter $\overline{A D}$ such that $|A B|=|B C|=1$. Calculate $|C D|$.
## First Solution.

Let $S$ be the center of the given circle, and $T$ the intersection of segments $\overline{S B}$ and $\overline{A C}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_12fcd97ef28da03e8b64g-02.jpg?height=549&width=854&top_left_y=565&top_left_x=607)

The angle $\varangle A C D$ is a right angle because it is an inscribed angle over the diameter of the circle.

Notice that $\varangle A S B = \varangle C S B$, so $\varangle A B S = \varangle C B S$. By the S-S-S congruence theorem, triangles $A B T$ and $C B T$ are congruent. In particular, $\varangle A T B = \varangle C T B$, so we conclude $\varangle A T B = 90^{\circ}$, i.e., $A T \perp B S$. In other words, $\overline{A T}$ is the altitude of triangle $A S B$.

Let $v_{a}$ be the altitude of triangle $A S B$ from vertex $B$. Since $|A B| = 1$, $|A S| = \frac{3}{2}$, and triangle $A S B$ is isosceles, we have

$$
v_{a} = \sqrt{\left(\frac{3}{2}\right)^{2} - \left(\frac{1}{2}\right)^{2}} = \sqrt{2}
$$

The length of $|A T|$ can be determined by expressing the area of triangle $A S B$ in two ways:

$$
\frac{|A T| \cdot |B S|}{2} = \frac{|A B| \cdot v_{a}}{2}
$$

From this, we get $|A T| = \frac{2 \sqrt{2}}{3}$, so $|A C| = 2|A T| = \frac{4 \sqrt{2}}{3}$, and finally, by the Pythagorean theorem,

$$
|C D| = \sqrt{|A D|^{2} - |A C|^{2}} = \frac{7}{3}
$$