# Problem regarding the intersection of a circumscribed circle and an exterior angle bisector and the midpoint of an arc

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:

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 $$\widehat{CAB}$$?

Note that $$H$$ is the midpoint of the arc $$\widehat{BHC}$$, since $$H$$ being on the bisector of $$\angle BAC$$ implies that inscribed angles over $$BH$$ and $$HC$$ are equal, so these arcs are equal too. Now, $$\angle HAE= 90^\circ$$ since interior and exterior angle bisectors are perpendicular, so $$EH$$ is the diameter of the circle. Now $$H$$ being the midpoint of the arc $$\widehat{BHC}$$ implies that $$E$$ is the midpoint of the arc $$\widehat{CEB}$$.
• Thank you very much! I totally missed the right-angle $\measuredangle HAE$, but it was so obvious! 😅 Thank you once again! – Invisible Jun 21 '20 at 15:12