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$PQRS$ is a cyclic quadrilateral. If $\overline{SQ}$ bisects $\angle PQR$ prove chord $\overline{PS}$ = chord $\overline{SR}$.

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    Here's a sketch: $\angle PQS$ and $\angle RQS$ would be congruent, and thus subtend congruent arcs. Since $\stackrel{\frown}{PS}$ and $\stackrel{\frown}{SR}$ are congruent, then...2011-08-04

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Note that The Inscribed Angle Theorem says that $\angle PQS$ is half the central angle of chord $\overline{PS}$ and $\angle SQR$ is half the central angle of chord $\overline{SR}$. Since $\overline{SQ}$ bisects $\angle PQR$, $\angle PQS = \angle SQR$. Therefore, $\overline{PS}=\overline{SR}$ by SAS (side-angle-side).

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    @MrK: Without first proving a theorem that says that the length of a chord is dependent only on the diameter of the circle and the cosine of the inscribed angle subtended by the chord, I don't see a way without introducing some addition construction.2011-08-04