$PQRS$ is a cyclic quadrilateral. If $\overline{SQ}$ bisects $\angle PQR$ prove chord $\overline{PS}$ = chord $\overline{SR}$.
Asked
Active
Viewed 1,492 times
2
-
Sheesh, I already embedded the image for you and cleaned up your notation... – J. M. ain't a mathematician Aug 04 '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 04 '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... – J. M. ain't a mathematician Aug 04 '11 at 13:16
1 Answers
2
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).
robjohn
- 345,667
-
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 04 '11 at 13:29
-
out of curiosity is is possible to prove this using the triangles on the diagram? – MrK Aug 04 '11 at 14:02
-
1@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 04 '11 at 16:40