Existence of Simsons Line using complex numbers Let $ABC$ be a triangle and $P$ is a point on its circumcircle. $X,Y$ and $Z$ are the feets of the perpendicular from $P$ on $BC, CA$ and $AB$, respectively. Prove that points $X, Y, Z$  are collinear.
Now proving this using angle chasing isn't hard. But I want to prove it using complex numbers.
Set $O=0$ where $O$ is the center of the circumcircle of $ABC$ ($(ABC)$ is the unit circle) and denote the corresponding complex number of every point with its lowercase letter. Here is the formula for each feet,
$$x=\frac{1}{2}(b+c+p-bc\bar p), \quad y= \frac{1}{2}( a+c+p-ac\bar p),\quad z= \frac{1}{2}( a+b+p-ab\bar p)$$
We know $x,y,z$ are collinear iff $x-y/x-z$ is real. But $$\frac{x-y}{x-z}=\frac{b-a-c\bar p(b-a)}{c-a-b\bar p(c-a)}=\frac{(b-a)(1-c\bar p)}{(c-a)(1-b\bar p)}$$ Is that even real?

Noting $\;\overline{p}=1/p \;$ we have
$$\frac{(b-a)(1-c\bar p)}{(c-a)(1-b\bar p)}=\frac{(b-a)(p-c)}{(c-a)(p-b)} =\frac{a-b}{a-c}\div \frac{p-b}{p-c}$$
which is real since $A,B,C,P \;$ are concylic. $\; \blacksquare$

Alternatively, it also suffices to verify
$$\frac{\left(\frac{1}{b}-\frac{1}{a}\right) \left(\frac{1}{p}-\frac{1}{c}\right)}{\left(\frac{1}{c}-\frac{1}{a}\right) \left(\frac{1}{p}-\frac{1}{b}\right)}= \cdots = \frac{(b-a)(p-c)}{(c-a)(p-b)}.$$
