$\underline{\textbf{Claim:}}$ $ QA=QB+QC$ $\underline{\textbf{Proof:}}$By ptolmey's theorem on ABCQ, we get $AB\cdot QC+AC \cdot BQ =BC \cdot AQ$ $\text{ Also as AB = BC =AC, the claim follows}$ So we need to show that $\frac{1}{QD} = \frac{1}{QB} + \frac{1}{QC}=\frac{QB+QC}{QB\cdot QC}=\frac{QA}{QB\cdot QC}\iff QA\cdot QD = QB \cdot QC$ Without the loss of generality , set $a=1, c = \omega,b=\omega^2$, where $\omega = \frac{-1+i\sqrt{3}}{2}$ Let $d = -0.5 + ix , x\in \mathbb{R}$ Now as $Q$ lies on $AD$, we get $\begin{vmatrix}1&1&1\\-0.5+ix&-0.5-ix&1\\q&\bar{q}&1\end{vmatrix} = 0\iff\begin{vmatrix}1&1&1\\-1&1&\frac{3i}{2x}\\q^2&1&q\end{vmatrix}=0$$\iff \left(\frac{3i}{2x}-1\right)q^2+2q-\left(1+\frac{3i}{2x}\right)=0$ $q=1$ is one of the roots, hence by vieta , the other root should be $q = \frac{2x+3i}{2x-3i} $ Now we need to show that $|q-a|^2|q-d|^2 = |q-b|^2|q-c|^2 \iff |q^2-(a+d)q+ad|^2=|q^2-(b+c)q+bc|^2$ We have \begin{align*}a+d &= 0.5 +ix\\ ad &= -0.5+ix\\ b+c &= -1\\ bc &= 1 \end{align*} So we need to show that $\left|q^2-\left(\frac{1}{2}+xi\right)q-\frac{1}{2}+xi\right|^2=\left|q^2+q+1\right|^2$ On observing, we see that $-\left(\frac{1}{2}+xi\right)q-\frac{1}{2}+xi = q+1$, hence we are done.

was just practising complex bash lel