$PQRS$ is a cyclic quadrilateral. If $\overline{SQ}$ bisects $\angle PQR$ prove chord $\overline{PS}$ = chord $\overline{SR}$.
[Math] PQRS is a cyclic quadrilateral. If SQ bisect
euclidean-geometry
Related Solutions
Why not just use the property of rectangles that diagonals of rectangle bisect each other?
That will mean that the intersection of diagonals is a point from which all the vertices are equidistant, which means, from the definition of a circle, that they lie on the circle with the center at the intersection.
If you had to prove the property first, you can do it by constructing the two diagonals, intersecting at O, then showing that any two vertically opposite triangles are congruent.
You don't need a nice diagram, a lousy one will do just as well
Let the angles of the original quadrilateral be $2\alpha,2\beta,2\gamma,2\delta$ so that $$\alpha+\beta+\gamma+\delta=180$$
So according to the diagram, the opposite angles in the small quadrilateral add to $180$
Best Answer
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).