In the very beginning, I'm going to refer to a very similar question where, unlike in my task, there is an assumption the intersection of the exterior angle bisector and a circumscribed circle is the midpoint of the arc.
$\triangle ABC$ is given where $|AB|>|AC|$. Bisector of the exterior angle $\measuredangle BAC$ intersects the circumscribed circle of $\triangle ABC$ at the point $E$. Point $F$ is the orthogonal projection of the point $E$ onto the line $AB$. Prove $|AF|=|FB|-|AC|$.
Attempt:
I adapted the answer by @Futurologist to fit my notation.
Going from the $E$ being the midpoint of the arc $\widehat{CAB}$, let $D\in BC$ s. t. $|AD|=|AC|$, $C\in\overline{BD}$, $\triangle DAC$ is isosceles. Now, $EA$ is the interior angle of $\measuredangle DAC$ (located on the $y$ axis in my picture, whereas the $x$-axis is the interior angle bisector of $\measuredangle BAC$).
Since $\triangle DAC$ is isosceles, $EA$ is also an orthogonal bisector of the edge $CD$. Let $P\equiv EA\cap CD$. Then $|DP|=|PC|$.
Since $E$ is the midpoint of $\widehat{CAB}$, $\color{red}{|EB|}=|EC|=\color{red}{|ED|}\implies\triangle DEB$ is isosceles and $\overline{EF}$ is its altitude $\implies |DF|=|FB|$. $$|FB|=|DF|=|DA|+|AF|=|AC|+|AF|\iff |AF|=|FB|-|AC|$$ Since the information that $E$ is the midpoint of the $\widehat{CAB}$ isn't given, I believe I have to prove it.
I know that:
$$\boxed{\measuredangle CAB=\measuredangle CEB}$$
and
$EF\perp AB\ \land\ EA\perp AH\implies\measuredangle AEF=\measuredangle HAB$, where $AH$ is the interior angle bisector of $\measuredangle BAC$.
If set the vertex $A$ to be at the origin, then the edges $\overline{AC}$ and $\overline{BC}$ belong to the lines $y_{1,2}=\pm k,k\in\Bbb R,$ but it doesn't look like progress.
May I ask for advice on how to prove $E$ is the midpoint of $\widehat{CAB}$?
Thank you in advance!
