<s>[INST] Given a triangle ABC inscribed in the unit circle , the 3 vertices could be described via 3 complex number, namely, $a$, $b$, and $c$.
Now $AD$ is an altitude, $D$ is the foot of $AD$ on $BC$.
Prove: $$D = \frac{a+b+c}2 - \frac{bc}{2a}$$
So far my progress is --

*

*the circumcentre of triangle $ABC$ is just $O = 0$.

*its centroid $G = \frac{a+b+c}3$.

*by Euler line, the orthocentre $H = a+b+c$.

*also, we can see that the centre of the nine point circle $N =
   \frac{a+b+c}2$.

But then I'm a bit stuck. $D$ is on the line $AH$ and $BC$, but I couldn't reach the conclusion to be proved.
 [/INST] line $AH$:
$$\frac{z-a}{b+c} = \overline{(\frac{{z-a}}{b+c})}$$
line $BC$:
$$\frac{z-b}{c-b} = \overline{(\frac{{z-b}}{c-b})}$$
so in the first equation:
$$\bar z = \overline{b+c} \cdot \frac{z-a}{b+c} + \bar a$$
in the second equation:
$$\bar z = \overline{c-b} \cdot \frac{z-b}{c-b} + \bar b$$
now $$a \bar a = b \bar b = c \bar c = 1$$
so:
$$\overline{c-b} \cdot \frac{z-b}{c-b} + \bar b = \overline{b+c} \cdot \frac{z-a}{b+c} + \bar a$$
$$(\frac1c-\frac1b) \cdot \frac{z-b}{c-b} + \frac1b = (\frac1b+\frac1c)\cdot \frac{z-a}{b+c} + \frac1 a$$
$$\frac{b-z}{bc} + \frac1b = \frac{z-a}{bc} + \frac1 a$$
$$\frac1c+ \frac1b -\frac1a= \frac{2z-a}{bc} $$
$$z = \frac{a+b+c}2 - \frac{bc}{2a}$$
our answers are a bit different. Did I miss something?
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