Ariel Wilkinson

2022-10-01

Proof with congruence of angles

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.

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.

odejicahfc

Beginner2022-10-02Added 10 answers

Step 1

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). $\mathrm{\angle}QPU=\mathrm{\angle}PQU=x$, and $\mathrm{\angle}QPV=\mathrm{\angle}PQV=x$, so $\mathrm{\angle}PUQ=\mathrm{\angle}PVQ$. 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.

Step 2

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.

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). $\mathrm{\angle}QPU=\mathrm{\angle}PQU=x$, and $\mathrm{\angle}QPV=\mathrm{\angle}PQV=x$, so $\mathrm{\angle}PUQ=\mathrm{\angle}PVQ$. 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.

Step 2

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.

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