## Task A-2.5.

Let $\overline{A B}$ be the diameter of circle $k$, and let point $O$ be its center. Let $C$ be a point outside circle $k$ on the perpendicular bisector of segment $\overline{A B}$. Segment $\overline{A C}$ intersects circle $k$ at point $D$. If $|A B|=2$ and $|C D|=1$, determine $|O C|$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9afe9556dd3f84037ce6g-11.jpg?height=865&width=691&top_left_y=1389&top_left_x=611)

Since $|OD| = |CD| = 1$ (the length of $\overline{OD}$ is the radius of circle $k$), triangle $OCD$ is isosceles, so $\angle COD = \angle DCO$.

Similarly, because $|OA| = |OD| = 1$, triangle $OAD$ is isosceles and $\angle OAD = \angle ODA$.

Since $\angle AOC = 90^\circ$, it follows that $\angle AOD = 90^\circ - \angle COD = 90^\circ - \angle DCO$.

On the other hand, from the right triangle $AOC$, we have $\angle OAC = 90^\circ - \angle DCO$, so $\angle OAD = \angle AOD$. In triangle $AOD$, all angles are equal, so it is equilateral. 1 point

Now we can conclude that $|AD| = 1$, and thus $|AC| = 2$. 1 point

According to the Pythagorean theorem for triangle $AOC$, we have $|OC| = \sqrt{|AC|^2 - |OA|^2} = \sqrt{3}$.