# 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!

• Such a nice diagram! I just recently found out about the software called "Geogeba". Is that what you used to make the diagram? – quasi Apr 18 '17 at 17:24
• @Olivia Ryan How did you end up with all integers for segments? It would not be by chance ... Maybe remaining sides are also integers? – Narasimham Apr 18 '17 at 19:14

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?

• Nice. I posted an answer (now deleted). I started the same way, but I missed the trick that one only needs to find the value of $x+y$, not the values of $x,y$. – quasi Apr 18 '17 at 17:06
• Thank you! but how did you get those equations? what properties did you use? – Parley Apr 18 '17 at 17:06
• @Olivia Ryan -- Power of a Point. – quasi Apr 18 '17 at 17:07
• @quasi for point S and point R? – Parley Apr 18 '17 at 17:08
• @Olivia Ryan -- en.wikipedia.org/wiki/Intersecting_chords_theorem – quasi Apr 18 '17 at 17:08

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 $27$.
• is there any way you can explain where to use Ptolemy? – Parley Apr 18 '17 at 17:23
• I didn't use Ptolemy's Theorem. Power of a Point is a natural choice for this question. It can be proved using similar triangles -- it's not hard. In a course that discusses cyclic quadrilaterals, I'm surprised it hasn't already been introduced. – quasi Apr 18 '17 at 17:26