How to prove Circle Angle Theorem using Vectors

Question

Please Let me know how to prove this theorem using vectors.

Theorem: The angle of the centre of circle is twice the angle at the circumference.

For example:

Answer 1

It is well known that all the inscribed angles $\angle CDB$ (vertex $D$ following your figure) have equal measure hence we can make $\overline{DC}$ or $\overline{DB}$ be a diameter. Let $r$ be the radius of the circle. Denote the area of the triangle of vertices $A,B,C$ by $\triangle ABC$ Clearly $$\triangle CDB=\triangle ADB+\triangle CAB$$ Now using vectors one has $$\triangle CDB=\big | \frac{\vec{DC}\text{ x }\vec{DB}}{2}\big|=\big | \frac{|\vec{DC}|.|\vec{DB}|\sin\beta}{2}\vec u_k\big|=r|\vec{DB}|\sin\beta$$ $$\triangle CAB=\big | \frac{\vec{AC}\text{ x }\vec{AB}}{2}\big|=\big | \frac{|\vec{AC}|.|\vec{AB}|\sin\alpha}{2}\vec u_k\big|=\frac{r^2\sin\alpha}{2}$$ $$\triangle ADB=\big | \frac{\vec{AD}\text{ x }\vec{AB}}{2}\big|=\big | \frac{|\vec{AD}|.|\vec{AB}|\sin (180^{\circ}-\alpha)}{2}\vec u_k\big|=\frac{r^2\sin\alpha}{2}$$ It follows $$ r|\vec{DB}|\sin\beta=r^2\sin \alpha$$ $$ r|\vec{DB}|\sin\beta=r^2\sin \alpha\iff \sin\alpha=\frac{|\vec{DB}|}{r}\sin \beta$$ Since the triangle $DBC$ is rectangle we have $$\frac{|\vec{DB}|}{r}=\frac{2|\vec{DB}|}{2r}=2\cos \beta$$ Thus we have finish because $\sin\alpha=2\sin\beta\cos\beta$ means that $$\color{red}{\alpha=2\beta}$$