A very different, difficult geometry problem that need to be helped- The problem:


  In $\triangle PQR$, $ST$ is the perpendicular bisector of $PR$, and $PS$ is the bisector of $\angle QPR$ and the point $S$ lies on $QR$. If $|QS|=9$ and $|SR|=7$ and $|PR|=x/y$, where $x$ and $y$ are co-primes, then what is $x+y$?


I can come up with these steps only:

$\angle STR$ is equal to $90^\circ$, so $\triangle STR$ is a right triangle. The same with $\Delta STP$. In $\triangle PST$ and $\triangle STR$, $ST\cong ST$, $PT\cong TR$, and $\angle PTS= \angle STR$, so that $\triangle PST\cong\triangle STR$. Therefore, $|PS|=|SR|=7$, and $\angle SPT\cong\angle SRT$.

That's the end.

I have no clue how to solve the next steps, or to solve the problem. I don't even know if the above steps lead toward an answer to the problem!

So please help me see how I can solve this step. I am a 8th grader. So please be careful, so that I can understand the steps.

Sorry, I can't add the figure.
$\angle SRP=\angle SPR = \angle SPQ=\theta,$ say.  Then adding up the angles in $\triangle PQR$ gives $\angle Q = 180^{\circ}-3\theta,$ and adding up the angles in $\triangle PSQ$ gives $\angle PSQ = 2\theta=\angle QPR.$  That is, $\triangle  PQR$ and $\triangle PSQ$ are similar.

Corresponding sides of similar triangles are proportional, so we have $$
\frac{|QP|}{9}=\frac{16}{|QP|}$$  (On the left-hand side we have two sides of $\triangle PQS,$ and on the right-hand side two we have two sides of $\triangle PQR.)$  We conclude that $|QP|=12.$  Now using similarity once again, we have $$\frac{|PR|}{|PQ|}=\frac{7}{9}\implies |PR|=\frac{28}{3}$$

That is, $\boxed{x=28,\space y=3,\space x+y=31}$.
I think this question is contradictory. If you draw a line segment $PR$ and perpendicularly bisect it. Now draw $angle QPR$ with Q as an unfixed point. You will see that the value of $PT,PS,$ $angle QPR$ will remain same but size of $QS$ will differ as $Q$ is an unfixed point along $rayPQ$.
So length of $QS$ is unwanted information.
Hence, it is impossible to find ${x\over y}$.