<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] 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.
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