Proof with congruence of angles I came across a proof exercise from my proof work-book that I am stuck on.  
The questions says:  
Suppose we have angle PQR with P, Q, and R non-collinear, and ray QS distinct from ray QR such that angle PQS is congruent to angle PQR.  Prove that if angle PQT is congruent to angle PQR, then either ray QT = ray QR or ray QT = ray QS.
From the question I was able to get that angle PQS is congruent to angle PQS is congruent to angle PQR.  I am not sure where to go from here or what theorems to use.
Any help would be greatly appreciated.
Thanks in advance,
Michael
 A: We'll assume given a line segment PQ, and three distinct rays QR, QS, and QT, each making the same angle $x$ with PQ; we'll show this leads to a contradiction. 
We'll assume $x$ is not a right angle; we'll come back to deal with that case later. 
There's a line L through P, parallel to QR. This line meets the ray QS at U, and it meets the ray QT at V (this is where we need the assumption that $x$ is not a right angle). $\angle QPU=\angle PQU=x$, and $\angle QPV=\angle PQV=x$, so $\angle PUQ=\angle PVQ$. So line segments QU and QV make the same angle with line L, so these line segments are parallel. But that's impossible, since they meet at Q. 
Now if $x$ is a right angle, then the line L through P parallel to QR can't meet either of the rays QS and QT --- if it did, you'd get a triangle with two right angles. So QR, QS, and QT are all rays through Q parallel to L. But that says there are (at least) two lines through Q parallel to L, which contradicts the parallel postulate, and we're done. 
