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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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Sheesh, I already embedded the image for you and cleaned up your notation... –  Guess who it is. Aug 4 '11 at 13:11
thanks, I'm inexperienced with the notation on web and as a new user I couldn't embed an image. –  MrK Aug 4 '11 at 13:13
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... –  Guess who it is. Aug 4 '11 at 13:16

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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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thanks robjohn, I can't believe I overlooked that. I was trying to use a congruent triangles proof, but couldn't get it to work. –  MrK Aug 4 '11 at 13:29
out of curiosity is is possible to prove this using the triangles on the diagram? –  MrK Aug 4 '11 at 14:02
@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. –  robjohn Aug 4 '11 at 16:40

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