Proof that angle 1 is congruent to angle 3 using triangle congruence and vertical angles.

The image presents a geometry proof involving a triangle and some angle relationships. Let's go through the given information, what needs to be proven, and then analyze the proof step by step.

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🔷 Given:

- $ \overline{RT} \cong \overline{ST} $

This means that sides $ RT $ and $ ST $ are congruent (equal in length).

🔷 Prove:

- $ \angle 1 \cong \angle 3 $

We need to show that angle 1 is congruent to angle 3.

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🔷 Diagram Analysis:

From the diagram:

- Triangle $ \triangle RTS $ has points $ R $, $ T $, and $ S $.
- A straight line extends from point $ R $ to the left, forming angle $ \angle 1 $ with the segment $ RT $.
- $ \angle 2 $ is the angle at vertex $ R $ inside triangle $ RTS $.
- $ \angle 3 $ is the angle at vertex $ S $ inside the triangle.

So:
- $ \angle 1 $ and $ \angle 2 $ are vertical angles formed by the intersection of the line extending from $ R $ and segment $ RT $ — actually, more accurately, they are adjacent angles on a straight line? Wait — let’s clarify.

Wait! Actually, looking closely:

- The line goes through $ R $, and $ RT $ is one side of the triangle.
- So $ \angle 1 $ and $ \angle 2 $ are adjacent angles that form a linear pair — but the statement says they are vertical angles?

That seems like a mistake.

Let’s reevaluate.

But in the proof table, it says:

> 3.) $ \angle 1 \cong \angle 2 $
> Reason: Vertical angles are congruent

But vertical angles are formed when two lines intersect. If only one line is drawn through $ R $, and $ RT $ is part of another line, then unless there’s a second line crossing it, vertical angles may not exist.

But wait — perhaps the line passing through $ R $ is extended beyond $ R $, and $ RT $ is going up to $ T $. Then, the line continues past $ R $, so we have two lines intersecting at $ R $: one is the horizontal line (with $ R $ on it), and the other is $ RT $. But that would make $ \angle 1 $ and $ \angle 2 $ adjacent angles on a straight line — not vertical.

Unless... the line from $ R $ is crossed by another line? No, the diagram shows a triangle $ RTS $, and a horizontal line passing through $ R $, extending to the left.

So the two rays at $ R $ are:
- The horizontal ray going left from $ R $
- The ray $ RT $ going up to $ T $

Then $ \angle 1 $ is between the leftward extension and $ RT $, and $ \angle 2 $ is between $ RT $ and the rightward direction (toward $ S $).

Wait — if $ R $ lies on a straight line, and $ RT $ goes up from $ R $, then:
- $ \angle 1 $ and $ \angle 2 $ are adjacent angles that together form a straight line — meaning they are supplementary, not necessarily vertical.

But the proof says:
> 3.) $ \angle 1 \cong \angle 2 $
> Reason: Vertical angles are congruent

This is only true if $ \angle 1 $ and $ \angle 2 $ are vertical angles, which requires two intersecting lines forming an "X" shape.

So for $ \angle 1 $ and $ \angle 2 $ to be vertical angles, there must be two lines intersecting at $ R $: one is the horizontal line, and the other is the line $ RT $, but unless $ RT $ is extended beyond $ R $, we don’t get vertical angles.

Ah! Possibly, the line $ RT $ is extended beyond $ R $, forming a straight line from $ T $ through $ R $ and continuing to the left. Then, the horizontal line intersects this extended line at $ R $, forming four angles.

In that case:
- $ \angle 1 $ and $ \angle 2 $ are vertical angles if $ \angle 1 $ is opposite $ \angle 2 $ across the intersection.

But in the diagram, $ \angle 1 $ is labeled outside the triangle, on the left, and $ \angle 2 $ is inside the triangle at $ R $.

So if the line $ TR $ is extended beyond $ R $, and the horizontal line passes through $ R $, then yes — the two lines intersect at $ R $, forming vertical angles.

So:
- $ \angle 1 $ and $ \angle 2 $ are not adjacent — they are opposite each other — so they are vertical angles.

Therefore, they are congruent because vertical angles are always congruent.

That makes sense.

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✔ Now, let’s walk through the proof:

| Statement | Reason |
|---------|--------|
| 1.) $ \overline{RT} \cong \overline{ST} $ | Given |
| 2.) $ \angle 2 \cong \angle 3 $ | Angles opposite congruent sides in a triangle are congruent |
| 3.) $ \angle 1 \cong \angle 2 $ | Vertical angles are congruent |
| 4.) $ \angle 1 \cong \angle 3 $ | Transitive property |

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🔍 Step-by-step Explanation:

1. Given: $ RT \cong ST $

This tells us that triangle $ RTS $ is isosceles with $ RT = ST $. Therefore, the base angles opposite these sides are congruent.

2. $ \angle 2 \cong \angle 3 $ — Because in triangle $ RTS $, sides $ RT $ and $ ST $ are congruent, so the angles opposite them must be congruent.

- $ \angle 2 $ is opposite side $ ST $
- $ \angle 3 $ is opposite side $ RT $

So yes, by the Isosceles Triangle Theorem: *If two sides of a triangle are congruent, then the angles opposite those sides are congruent.*

3. $ \angle 1 \cong \angle 2 $ — These are vertical angles formed at point $ R $ where two lines intersect:
- One line: the horizontal line through $ R $
- The other line: the line $ TR $ extended beyond $ R $

Since vertical angles are always congruent, this step is valid.

4. $ \angle 1 \cong \angle 3 $ — By the Transitive Property of Congruence:
- If $ \angle 1 \cong \angle 2 $ and $ \angle 2 \cong \angle 3 $, then $ \angle 1 \cong \angle 3 $

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✔ Final Answer:

Yes, the proof is correct.

We are given that $ RT \cong ST $, so triangle $ RTS $ is isosceles, leading to $ \angle 2 \cong \angle 3 $. Then, since $ \angle 1 $ and $ \angle 2 $ are vertical angles, they are congruent. Finally, using transitivity, $ \angle 1 \cong \angle 3 $, which is what we wanted to prove.

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🟩 Conclusion:

✔ The proof is logically sound and correctly uses:
- Isosceles Triangle Theorem
- Vertical Angles Theorem
- Transitive Property

Thus, $ \angle 1 \cong \angle 3 $ is proven.