In the figure, two circles intersect at $P$ and $Q$... In the figure, two circles intersect at $P $ and $Q$. $O$ is the centre of the smaller circle which lies on the circumference of the larger circle and $RO$ is joined and produced to meet $QS$ at $X$. Prove that $RQ=RS$

My Attempt 
$1. \angle QSP=\frac {1}{2} arc QP$
Let $RQ$ intersects the circumference of smaller circle at $A$
$2. \angle AQS=\frac {1}{2} arc APS$.
I don't have any idea to move further. Please help.
Thanks. 
 A: Let K be the center and OKR’ the diameter of the red circle. Suppose that M is the intersecting point of KO and PQ.

From the fact that OK is the line of centers and PQ is the common chord, we can say that (1) $\angle KMP = \angle KMQ = 90^0$; and (2) PM = PQ. Including the common side MR’, we have $\triangle R’MP \cong \triangle R’MO$. This further means $\alpha = \beta$.
By angles in the same segment, all the green marked angles are equal.
By exterior angle of cyclic quadrilateral, $\phi = \theta$.
Then, $\omega = \phi + \epsilon = \theta + \lambda = \dfrac {1}{2} (\omega + \omega’)$ [Anlge at center = twice angle at circumference.]
∴ $\omega = \omega’$.
Hence, $ROX \bot QS$. Result follows.
A: Construction: Join OP and OQ

*

*chord OP = chord OQ [Radii of circle]


*arc OP = arc OQ [from statement 1, equal arcs subtended from equal
                            chords]



*<QRX = <SRX  [Equal inscribed angles subtended from equal arcs]


*<POQ = 2 <PSQ [Relation of inscribed and central angles]


*<POQ + <QRS = 180 [Sum of opposite angles of cyclic quad is
                                   supplementary]



*2<PSQ + <QRS = 180 [From statement 4 and 5]
or, <QRS = 180 - 2<PSQ


*<QRS + <PSQ + <RQS = 180 [Sum of angles of triangle RQS]


*180 - 2<PSQ + <PSQ + <RQS = 180 [From statement 6 and 7]


*<PSQ = <RQS [from statement 7]


*RQS is isosceles triangle [from statement 9, base angles of isosceles
                                         triangle is equal]



*RQ = RS  [Base sides of isoscels triangle are equal]
