Geometry of Complex Numbers (lemma)

Question

Recently I found very interesting lemma, unfortunately without proof. I will be grateful if you help me to prove it.
Let triangle ABC have incenter $I$ and circumcircle $Γ$. Lines $AI, BI, CI$ meet $Γ$ again at $D, E, F$ . If $Γ$ is the unit circle of the complex plane then there exists $x, y, z ∈ C$ satisfying $a = x^2, b = y^2, c = z^2$ and $d = −yz, e = −zx, f = −xy$.

Answer 1

To be sure that I get it, check me?
Since $\angle CBD$ and $\angle BCD$ based on equal arcs, then $BD= CD$. Thus \begin{align*} (b-d)(\overline{b-d}) &= (c-d)(\overline{c-d})\\ \text{,by circle equation, we get } \bar b = \frac{1}{b},\text{ } \bar{c}=\frac{1}{c}\text{ and } \bar d = \frac{1}{d}.\\ (b-d)(\frac{1}{b}-\frac{1}{d}) &= (c-d)(\frac{1}{c}-\frac{1}{d})\\ \frac{b}{d}+\frac{d}{b}&= \frac{c}{d}+\frac{d}{c}\\ c(b^2+d^2) &= b(c^2+d^2)\\ d^2 &= bc \end{align*} Analogously we get $e^2 = ca$ and $f^2 = ab$.
We pick an arbitrary choice for $x$ first. Then, of the two choices of $y$, we pick the one such that $-xy = f$. Similarly, for the two choices of $z$, we pick the one such that $-xz = e$. Since $AD\perp EF$, we get \begin{align*} \overrightarrow{\rm AD }\cdot \overrightarrow{\rm EF} &= \frac{1}{2}[(a-d)(\overline{e-f})+(\overline{a-d})(e-f)\\ &=\frac{1}{2}[(x^2-d)(\overline{zx-xy}+(\overline{x^2-d})(zx-xy)]\\ &=\frac{1}{2}[(x^2-d)(\frac{1}{zx}-\frac{1}{xy})+(\frac{1}{x^2}-\frac{1}{d})(zx-xy)]\\ &=\frac{(x^2-d)(y-z)}{2x}(\frac{1}{yz}+\frac{1}{x}) \end{align*} Which gives us $d = -yz$.