In a circle with a radius of $10 \mathrm{~cm}$, segment $A B$ is a diameter and segment $A C$ is a chord of $12 \mathrm{~cm}$. Determine the distance between points $B$ and $C$.
Given that $AB$ is a diameter, the triangle $\triangle ABC$ is inscribed in a semicircle, implying that this triangle is a right triangle at vertex $C$. By the Pythagorean Theorem,

$$
BC^{2}=AB^{2}-AC^{2}
$$

that is,

$$
BC^{2}=20^{2}-12^{2}=256=16^{2}
$$

Thus, we obtain that $BC=16$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-112.jpg?height=348&width=420&top_left_y=1528&top_left_x=1412)