Q:

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. 

A:

$\angle SRP=\angle SPR = \angle SPQ=\theta,$ 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, $\boxed{x=28,\space y=3,\space x+y=31}$.

