I am also confused on Example 5 (~1:27:40) and how we got \( \frac{DM \cdot \sin (\angle PDM)}{DI \cdot \sin (\angle PDI)} = \frac{2R \cdot \sin^2 (\alpha)}{2R \cdot \sin (\alpha)} \cdot \frac{\sin(90- \alpha + \alpha - \beta)}{\sin(C + \alpha - \beta)} \)