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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.

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    $\begingroup$ Hint: triangles $BAC$ and $DEC$ have $AB=DE$ by construction, but that's not the only thing in common. $\endgroup$
    – dxiv
    May 18 '21 at 19:18
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    $\begingroup$ @dxiv Yes, angle CDE and ABC are congruent, but what does that lead to? $\endgroup$
    – Student
    May 18 '21 at 19:23
  • $\begingroup$ You haven't used "AC bisects the angle BAD" yet. That tells you something about point $C$. $\endgroup$
    – dxiv
    May 18 '21 at 19:24
  • $\begingroup$ @dxiv I see the obvious angles BAC and DAC are congruent. But what does C have to do with it? $\endgroup$
    – Student
    May 18 '21 at 19:29
  • $\begingroup$ Equal inscribed angles means equal subtended arcs, so $C$ is the midpoint of arc $BD$. $\endgroup$
    – dxiv
    May 18 '21 at 19:33
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  • It is given that the two red line segments are congruent.
  • $\angle ABC$ 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.
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