In a circle with radius $10 \mathrm{~cm}$, 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. This implies that this triangle is a right triangle at vertex $C$. By the Pythagorean Theorem,

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

that is,

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

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-27.jpg?height=352&width=415&top_left_y=526&top_left_x=1321)

Thus, we obtain that $BC=16$.