A proposition of trigonometry, or complex numbers

Prop. If $0,1,z,w$ all lie on a circle centered at $1/2$ then $zw$ is the orthogonal projection of $0$ onto the line through $z$ and $w$.

Proof. If $z,w$ to lie on that circle, then $$(z-\frac12)(\overline{z}-\frac12)=\frac14,\text{ which simplifies to }z\overline{z}=\frac{z+\overline{z}}2$$ $$(w-\frac12)(\overline{w}-\frac12)=\frac14,\text{ which simplifies to }w\overline{w}=\frac{w+\overline{w}}2.$$ Introducing names $z=a+bi,w=c+di$ these equations turn into $$a^2+b^2=a,\quad c^2+d^2=c.$$ Two things need to be shown: $(1)$ $zw$ lies on the line through $z$ and $w$, and $(2)$ $zw$ is perpendicular to $z-w$.

$(1)$ Verify that the following determinant is zero. $$\begin{align*} \begin{vmatrix}1&a&b\\1&c&d\\1&ac-bd&ad+bc\end{vmatrix}&=acd+bc^2-acd+bd^2-a^2d-abc+abc-b^2d+ad-bc\\ &=b(c^2+d^2)-(a^2+b^2)d+ad-bc=bc-ad+ad-bc=0. \end{align*}$$ $(2)$ Verify that $zw/(z-w)$ is imaginary, or equivalently that $zw(\overline z-\overline w)$ is imaginary. $$\begin{align*} zw(\overline z-\overline w)+\overline{zw(\overline z-\overline w)}&=z\overline zw-zw\overline w+z\overline z\overline w-\overline zw\overline w=z\overline z(w+\overline w)-(z+\overline z)w\overline w\\ &=\frac{z+\overline z}2(w+\overline w)-(z+\overline z)\frac{w+\overline w}2=\frac12(z+\overline z)(w+\overline w)(1-1)=0. \end{align*}$$