Proving a perpendicular bisector in a circle The quadrilateral $$ABCD$$ is inscribed in a circle, $$\overline{AC}$$ bisects $$\angle BAD$$, $$\overline{AD}$$ is extended to $$E$$ such that $$DE = AB$$. Prove that $$C$$ is on the perpendicular bisector of $$\overline{AE}$$.

This relates to circles and angles of inscribed quadrilaterals. Any help or advice is appreciated. The perpendicular bisector is drawn, though gray.

• Hint: triangles $BAC$ and $DEC$ have $AB=DE$ by construction, but that's not the only thing in common.
– dxiv
May 18 '21 at 19:18
• @dxiv Yes, angle CDE and ABC are congruent, but what does that lead to? May 18 '21 at 19:23
• You haven't used "AC bisects the angle BAD" yet. That tells you something about point $C$.
– dxiv
May 18 '21 at 19:24
• @dxiv I see the obvious angles BAC and DAC are congruent. But what does C have to do with it? May 18 '21 at 19:29
• Equal inscribed angles means equal subtended arcs, so $C$ is the midpoint of arc $BD$.
– dxiv
May 18 '21 at 19:33 • $$\angle ABC$$ and $$\angle ADC$$ are supplementary, since they are opposite angles of a cyclic quadrilateral, so the two marked angles are congruent.
• It is given that $$\angle BAC\cong\angle CAD$$, so arcs $$BC$$ and $$CD$$ are congruent. Therefore, the two green line segments, which cap those congruent arcs must also be congruent.
• Therefore, by SAS, the two triangles are congruent. Thus, by CPCTC, we can conclude that $$\overline{AC}\cong\overline{EC}$$. Since the perpendicular bisector of $$\overline{AE}$$ is the locus of points equidistant from $$A$$ and $$E$$, we have shown that $$C$$ must lie on that line.