Proof with congruence of angles

Question

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

Angle ABC makes one appearance in this question, and then is never heard from again. Was it supposed to be something else? — Gerry Myerson, Oct 21 '13 at 12:51
@GerryMyerson I've edited the question, could you give a hint? — Michael Ferashire Silva, Oct 21 '13 at 12:53
@GerryMyerson Thanks a lot, I understand the proof you have given :). I'd give it an up-vote, but I don't have enough reputation. — Michael Ferashire Silva, Oct 24 '13 at 16:05

score 0 · Accepted Answer · answered Oct 22 '13 at 12:07

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.

Proof with congruence of angles

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