Cyclic Quadrilateral and Ptolemy to find the length of a segment $ABCD$ is a cyclic quadrilateral with $AB=11$ and $CD= 19$. Points $P$ and $Q$ are on $AB$ and $CD$ respectively such that $AP=6,\, BP=5\, DQ=7$ and $CQ=12,\, PQ=27$. Extend $PQ$ till it  meets the circle at point $R$ and point $S$.
Find $RS$. 

This is what I have so far:


*

*I constructed chords $DS,\, CS,\, AD,\, BC,\, AR,\, RB$ and $BC$ 

*Ptolemy in ABCD   $AB\times DC+ AD \times BC= AC \times BD$

*I think Ptolemy needs to be done multiple times in all the cyclic quads 
i just don't know where to start or how to do that 


Thank you so much! 
 A: Ptolemy's theorem is not really needed to solve the problem.
Let $PR=x$ and $SQ=y$. We have $7\cdot 12=y(27+x)$ and $5\cdot 6=x(27+y)$.
Can you guess why and what $x+y+27$ is, as a consequence?
A: No need for Ptolemy's Theorem.

Instead, use Power of a Point . . .

Let $x = RP$ and $y = QS$.

Applying Power of a Point to the intersecting chords $AB$, $RS$, we get
$$(x)(y+27) = (5)(6)$$
Applying Power of a Point to the intersecting chords $CD$, $RS$, we get
$$(y)(x + 27) = (7)(12)$$
So now you have two equations in two unknowns. I'll leave the rest as an outline ...


*

*Subtract the equations to get a linear equation.$\\[4pt]$

*Solve for one of the unknowns in terms of the other.$\\[4pt]$

*Substitute the result into one of the original equations, thus getting a quadratic equation in one unknown.$\\[4pt]$

*By symmetry, $x,y$ are both roots of the same quadratic.$\\[4pt]$

*Since you only need the value of $x+y$ (thanks to Jack D'Aurizio for that observation), just use Vieta's formula to get the sum of the roots.$\\[4pt]$

*Finally, add $27$.

