Derivation of formula for Circumcenter of Triangle

Question

Given three complex numbers $z_{1},z_{2},z_{3}$ which are represented on the Argand plane by affixes $A,B,C$ respectively.

A diameter of the circumcircle of $\triangle ABC$ passing through $A$ is drawn which meets the triangle at $D$.

Can it be shown that $D$ divides $BC$ in the ratio $\sin 2C:\sin 2B$.

I was trying to prove that the circumcenter of $\triangle ABC$ is

$\frac{z_{1}\sin 2A+z_{2}\sin 2B+z_{3}\sin 2C}{\sin 2A+\sin 2B+\sin 2C}$ for which I require the ratio $BD:DC$

Answer 1

$$\frac{BD}{DC}=\frac{A(ABD\triangle)}{A(ACD\triangle)}=\frac{AB\cdot AD\cdot \sin(90^{\circ}-C)}{AC\cdot AD\cdot \sin(90^{\circ}-B)}=\frac{AB}{AC}\cdot\frac{\cos C}{\cos B}=\frac{\sin C}{\sin B}\cdot\frac{\cos C}{\cos B}=\frac{2\sin C \cos C}{2\sin B\cos B}=\frac{\sin 2C}{\sin 2B},$$ where for example $A(ABD\triangle)$ means the area of $ABD\triangle$.