Complex bash with $ABC$ inscribed in the unit circle so that \begin{align*} |a|=|b|=|c|&=1 \\ e &= \frac{a+c}2 \\ f &= \frac{a+b}2 \\ k = c + e - bc\overline{e} &= \frac{3ac+a^2-ab-bc}{2a} \\ \ell = b + e - bc\overline{f} &= \frac{3ab+a^2-ac-bc}{2a} \end{align*}Then \begin{align*} a-k &= \frac{a^2+ab+bc-3ac}{2a} \\ a-\ell &= \frac{a^2+ac+bc-3ab}{2a} \end{align*}and so since $a-\ell = bc(\overline{a}-\overline{k})$, we have $|a-k| = |a-\ell|$. $\blacksquare$