Proving a perpendicular bisector in a circle
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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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$\begingroup$- 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.
