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.
 A: 
I think this should work for all triangles.
 If you have a complex number $z=a+bi$, then transform it into $\vec z= a\hat i + b\hat j$.
 In the figure, we can write, the straight line of which BC is a part, as $$\vec r=\vec b+\lambda(\vec b -\vec c)$$ Now, let $$\vec d =\vec b+\lambda_1(\vec b -\vec c)$$. We have, by condition of perpendicularity, $(\vec d- \vec a) \cdot (\vec c - \vec b)=0$. Replacing $\vec d$, we have $(\vec b+\lambda_1(\vec b -\vec c) - \vec a) \cdot (\vec c - \vec b)=0$
 From which, $$\lambda_1=\frac{(\vec a-\vec c)\cdot(\vec b -\vec c)}{| \vec b -\vec c |^2} $$
so $$\vec d =\vec b+ \frac{(\vec a-\vec c)\cdot(\vec b -\vec c)}{| \vec b -\vec c |^2} (\vec b -\vec c)$$
 so if in general, $\vec z= z_1\hat i + z_2\hat j$, then
 $d_1\hat i + d_2\hat j = b_1\hat i + b_2\hat j + \frac{(a_1-c_1)(b_1-c_1)+ (a_2-c_2)(b_2-c_2)}{\sqrt{(b_1-c_1)^2+ (b_2-c_2)^2}} ((b_1-c_1)\hat i+ (b_2-c_2)\hat j)$
Now you can convert it back to complex numbers.
A: 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?
A: This is a solution for acute $\Delta ABC$ only
You could consider extending $AD$ such that it intersects the unit circle at $E$. It is not hard to see that in the situation where two chords intersect orthogonally, arc measures opposite the point of intersection add to $\pi$. Let $B$ sit $\beta$ from $A$ counterclockwise, and $C$ sit $\gamma$ from A clockwise, such that $E$ is $\pi - \gamma + \beta$ from $A$. Therefore, $E$ is given by the complex number $-\frac{bc}{a}$
In this case, then, $D$ is intersection point of the diagonals of cyclic quadrilateral $ABEC$. We now may use the Pythagorean theorem to arrive at the complex number representation of $D$.
Let the lengths $AD$, $BD$, $ED$, $CD$, be given by the letters $w, x, y, z$, and let the lengths $AB$, $BE$, $EC$, $CA$ be given by the letters $p, q, r, s$.
Then, by power of a point, we know $w^2y^2 = x^2z^2$. By applying Pythagoras:
$$(p^2-x^2)y^2 = x^2z^2$$
$$p^2y^2=r^2x^2$$
$$p^2(q^2-x^2) = r^2x^2$$
$$p^2q^2 = (p^2+r^2)x^2$$
$$x = \frac{pq}{\sqrt{p^2+r^2}}$$
Since $x$ is simply $BD$, and since $p = |b-a|, q = |\frac{bc}{a} + b|, r = |\frac{bc}{a}+c|$, then the complex number representation of $D$, call it $d$, can be given as:
$$d = b + \frac{(c-b)|b-a||b+\frac{bc}{a}|}{|c-b|\sqrt{|b-a|^2+|c+\frac{bc}{a}|^2}}$$
A: First of all determine the two complex numbers (vectors)
$$
\eqalign{
  & z = x + iy = b - a \to {\bf z} = (x,y)  \cr 
  & w = u + iv = c - a \to {\bf w} = (u,v) \cr} 
$$
Now, given two vectors, you can always decompose one of them
into a component parallel  and a component orthogonal to the other.
$$
{\bf z} = {\bf z}_{//{\bf w}}  + {\bf z}_{ \bot {\bf w}} \quad 
 \Rightarrow \quad \left\| {\bf z} \right\|^2  = \left\| {{\bf z}_{//{\bf w}} } \right\|^2
  + \left\| {{\bf z}_{ \bot {\bf w}} } \right\|^2 
$$
which is what you are looking for.
Clearly the parallel component is given by the dot product with the unitary $\bf w$
multiplied by the same
$$
{\bf z}_{//{\bf w}}  = \left( {{\bf z} \cdot {{\bf w} \over {\left\| {\bf w} \right\|}}} \right)
{{\bf w} \over {\left\| {\bf w} \right\|}}
$$
and for the orthogonal component by
$$
{\bf z}_{ \bot {\bf w}}  = {\bf z} - {\bf z}_{//{\bf w}} 
$$
and also consider the cross product
$$
{\bf z} \times {{\bf w} \over {\left\| {\bf w} \right\|}}
 = \left( {{\bf z}_{//{\bf w}}  + {\bf z}_{ \bot {\bf w}} } \right)
 \times {{\bf w} \over {\left\| {\bf w} \right\|}} = {\bf z}_{ \bot {\bf w}} 
 \times {{\bf w} \over {\left\| {\bf w} \right\|}}
$$
Then if you want to remain in the field of complex numbers, consider that
$$
\eqalign{
  & z\,w = \left( {xu - yv} \right) + i\left( {xv + yu} \right)  \cr 
  & \tilde z\,w = \left( {xu + yv} \right) + i\left( {xv - yu} \right) = \left( {{\bf z}
 \cdot {\bf w}} \right) + i\left\| {{\bf z} \times {\bf w}} \right\|  \cr 
  & \tilde z\,{w \over {\left| w \right|}} = \tilde z\,{w \over {\sqrt {\tilde ww} }}
 = \left( {{\bf z} \cdot {{\bf w} \over {\left\| {\bf w} \right\|}}} \right)
 + i\left\| {{\bf z} \times {{\bf w} \over {\left\| {\bf w} \right\|}}} \right\|
 = \left\| {{\bf z}_{//{\bf w}} } \right\| + i\left\| {{\bf z}_{ \bot {\bf w}} } \right\| \cr} 
$$
.. and the rest is obvious.
A: Instead of treating the points as arbitrary $\mathbb{R}^2$ vectors, let's make use of them being complex numbers.  If $z_1, z_2 \in \mathbb{C}$ are interpreted as vectors, then:

*

*They are perpendicular if the quotient between them is a pure imaginary number.

*They are parallel (or overlapping) if the quotient between them is a real number.

So,
$$AD \perp BC \implies \Re(\frac{d - a}{c - b}) = 0$$
$$BD \parallel BC \implies \Im(\frac{d - b}{c - b}) = 0$$
Also recall that $\Re(z) = \frac{z + \overline{z}}{2}$, $\Im(z) = \frac{z - \overline{z}}{2i}$, and conjugation is distributive over all of the four basic operations.  Therefore, from the first equation:
$$\Re(\frac{d - a}{c - b}) = 0$$
$$\frac{d - a}{c - b} + \frac{\overline{d} - \overline{a}}{\overline{c} - \overline{b}} = 0$$
$$(d - a)(\overline{c} - \overline{b}) + (\overline{d} - \overline{a})(c - b) = 0$$
$$\overline{c}d - \overline{b}d - a\overline{c} + a\overline{b} + c\overline{d} - b\overline{d} - \overline{a}c + \overline{a}b = 0$$
$$(\overline{c} - \overline{b})d + (c - b)\overline{d} = a\overline{c} + \overline{a}c - a\overline{b} - \overline{a}b$$
And from the other equation:
$$\Im(\frac{d - b}{c - b}) = 0$$
$$\frac{d - b}{c - b} - \frac{\overline{d} - \overline{b}}{\overline{c} - \overline{b}} = 0$$
$$(d - b)(\overline{c} - \overline{b}) - (\overline{d} - \overline{b})(c - b) = 0$$
$$\overline{c}d - \overline{b}d - b\overline{c} + b\overline{b} - c\overline{d} + b\overline{d} + \overline{b}c - b\overline{b} = 0$$
$$(\overline{c} - \overline{b})d + (b - c)\overline{d} = b\overline{c} - \overline{b}{c}$$
Adding these two equations together gives us:
$$2(\overline{c} - \overline{b})d = a\overline{c} + \overline{a}c - a\overline{b} - \overline{a}b + b\overline{c} - \overline{b}{c}$$
Now, to get rid of the conjugation operators.  Maybe there's something clever we can multiply everything by.  Hmm...
Ooh, I've got an idea!  If $z$ is on the unit circle, then $z\overline{z} = |z|^2 = 1$.  So let's just multiply everything by $abc$, and all of the conjugates will have something to cancel with.
$$2a(b - c)d = a^2b + bc^2 - a^2c - b^2c + ab^2 - ac^2$$
$$d = \frac{a^2b + bc^2 - a^2c - b^2c + ab^2 - ac^2}{2a(b - c)}$$
$$d = \frac{a^2(b - c) - bc(b - c) + a(b^2 - c^2)}{2a(b - c)}$$
$$d = \frac{a^2 - bc + a(b + c)}{2a}$$
$$d = \frac{a^2 + ab + ac}{2a} - \frac{bc}{2a}$$
$$d = \frac{a + b + c}{2} - \frac{bc}{2a}$$
Q.E.D.
A: just as my notes in case if anyone is interested.
Denote $D$'s coordinate as complex number $d$. Extend $AD$ until cross with the unit circle at point $X$, denote its coordinate as $x$.
First as $AX \bot BC$, we have
$$ax+bc=0$$
or $$x = -\frac{bc}{a}$$
Then as $\Delta ABD \sim \Delta CXD$, one has
$$\frac{DA}{DB}= -\frac{DC}{DX}$$
or
$$\frac{d-a}{d-b}= -\frac{d-c}{d-(-\frac{bc}{a})}$$
Assume $d\ne 0$, we get
$$d = \frac{a+b+c}2 - \frac{bc}{2a}$$
