Geometry Name: Proof Worksheet (3) Date: 1. Given: ∠3 ≅ ∠4 ... - Free Printable

Let’s solve each of the four geometry proofs step-by-step, using standard geometric reasoning (linear pairs, vertical angles, supplementary angles, etc.).

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Problem 1

Given: ∠3 ≅ ∠4
Prove: ∠1 ≅ ∠2

Diagram Notes:
Points A and B lie on a straight line. Triangle ABC is above the line AB.
- At point A: ∠1 and ∠3 are adjacent angles forming a linear pair → ∠1 + ∠3 = 180°
- At point B: ∠2 and ∠4 are adjacent angles forming a linear pair → ∠2 + ∠4 = 180°

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Proof:

| Statement | Reason |
|----------|--------|
| 1. ∠3 ≅ ∠4 | Given |
| 2. ∠1 + ∠3 = 180° | Linear Pair Postulate (angles at point A) |
| 3. ∠2 + ∠4 = 180° | Linear Pair Postulate (angles at point B) |
| 4. ∠1 = 180° - ∠3 | Subtraction Property of Equality (from 2) |
| 5. ∠2 = 180° - ∠4 | Subtraction Property of Equality (from 3) |
| 6. ∠1 = 180° - ∠4 | Substitute ∠3 for ∠4 (since ∠3 ≅ ∠4) |
| 7. ∠1 = ∠2 | Transitive Property (both equal to 180° - ∠4) |
| 8. ∠1 ≅ ∠2 | Definition of Congruent Angles |

✔ Conclusion: ∠1 ≅ ∠2

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Problem 2

Given: ∠1 ≅ ∠4
Prove: ∠2 ≅ ∠3

Diagram Notes:
Two lines intersecting, forming an “X” shape with angles labeled 1, 2, 3, 4 around the intersection.

Actually, looking closely — this looks like two transversals crossing, but more likely it's just two lines intersecting, forming vertical angles.

But given the labeling:
∠1 and ∠2 are adjacent, ∠2 and ∠3 are adjacent, etc. — this appears to be two intersecting lines, so:

- ∠1 and ∠3 are vertical angles → ∠1 ≅ ∠3
- ∠2 and ∠4 are vertical angles → ∠2 ≅ ∠4
- Also, ∠1 and ∠2 are a linear pair → ∠1 + ∠2 = 180°

Wait — but the given is ∠1 ≅ ∠4, and we’re to prove ∠2 ≅ ∠3.

In intersecting lines:
- Vertical angles are congruent: ∠1 ≅ ∠3, ∠2 ≅ ∠4
- So if ∠1 ≅ ∠4, then by transitivity: ∠1 ≅ ∠4 ≅ ∠2 → so ∠1 ≅ ∠2? That would only happen if they’re both 90°.

Actually, let’s use algebra.

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Proof:

| Statement | Reason |
|----------|--------|
| 1. ∠1 ≅ ∠4 | Given |
| 2. ∠1 + ∠2 = 180° | Linear Pair Postulate (adjacent angles on straight line) |
| 3. ∠3 + ∠4 = 180° | Linear Pair Postulate (adjacent angles on other straight line) |
| 4. ∠2 = 180° - ∠1 | Subtraction (from 2) |
| 5. ∠3 = 180° - ∠4 | Subtraction (from 3) |
| 6. ∠2 = 180° - ∠4 | Substitute ∠1 for ∠4 (since ∠1 ≅ ∠4) |
| 7. ∠2 = ∠3 | Transitive Property (both equal 180° - ∠4) |
| 8. ∠2 ≅ ∠3 | Definition of Congruent Angles |

✔ Conclusion: ∠2 ≅ ∠3

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Problem 3

Given: ∠1 ≅ ∠3
Prove: ∠2 is supplementary to ∠3

Diagram Notes:
Looks like two rays from point A: one going left-up (to C), one right-down (to T). Another ray goes up-right (angle 2), and another ray goes down-right (angle 3).
Actually — it’s probably three rays from point A:
- One ray forms ∠1 with the horizontal (left side)
- Another ray forms ∠2 (upward)
- Another ray forms ∠3 (right side)

Wait — better interpretation: Point A has three rays: AC, AT, and another ray going upward. The angles are:

- ∠1 between AC and the upward ray
- ∠2 between upward ray and AT
- ∠3 is the angle at point T? No — label says “∠3” near point T, but that doesn’t make sense unless it’s part of triangle or something.

Actually — re-examining: It seems like ∠1 and ∠2 are adjacent angles at point A, and ∠3 is an angle at point T, which is connected via segment AT.

Wait — perhaps it’s a diagram where:

- Point A has rays to C and T, and another ray in between.
- ∠1 is between CA and the middle ray.
- ∠2 is between middle ray and AT.
- Then ∠1 + ∠2 = ∠CAT (the whole angle).
- And ∠3 is at point T — maybe in triangle CAT?

But the problem says “∠2 is supplementary to ∠3” — meaning ∠2 + ∠3 = 180°.

Given: ∠1 ≅ ∠3

If we assume that points C, A, T form a triangle, and the middle ray creates ∠1 and ∠2 such that ∠1 + ∠2 = ∠CAT.

But without more info, let’s assume the simplest interpretation:

> The diagram shows two angles (∠1 and ∠2) that together form a straight angle (i.e., they are adjacent and sum to 180°), and ∠3 is congruent to ∠1. We are to prove ∠2 is supplementary to ∠3.

That makes sense.

So:

Assume ∠1 and ∠2 are adjacent and form a linear pair → ∠1 + ∠2 = 180°

Given: ∠1 ≅ ∠3 → ∠1 = ∠3

Then: ∠3 + ∠2 = 180° → ∠2 and ∠3 are supplementary.

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Proof:

| Statement | Reason |
|----------|--------|
| 1. ∠1 ≅ ∠3 | Given |
| 2. ∠1 + ∠2 = 180° | Linear Pair Postulate (assumed from diagram — ∠1 and ∠2 are adjacent on a straight line) |
| 3. ∠3 + ∠2 = 180° | Substitute ∠1 for ∠3 (since ∠1 ≅ ∠3) |
| 4. ∠2 is supplementary to ∠3 | Definition of Supplementary Angles |

✔ Conclusion: ∠2 is supplementary to ∠3

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Problem 4

Given: ∠4 ≅ ∠6
Prove: ∠5 ≅ ∠6

Diagram Notes:
Two parallel horizontal lines cut by a transversal. Angles labeled:

- ∠4 and ∠5 are at the bottom intersection (on the lower line)
- ∠6 is at the top intersection (on the upper line), same side as ∠4

Specifically:

- ∠4 and ∠5 are vertical angles? Or adjacent?
Looking at typical labeling:
At bottom intersection: ∠4 and ∠5 are adjacent — likely ∠4 is on the left, ∠5 on the right, forming a linear pair.

At top intersection: ∠6 is on the right side, same side as ∠4? Actually, if it’s a transversal cutting two lines:

Standard labeling:

- ∠4 and ∠6 are corresponding angles? But given ∠4 ≅ ∠6 — that would imply the lines are parallel.

But we’re not told the lines are parallel — we have to work with what’s given.

Actually — look at the diagram:
The transversal crosses two horizontal lines.

At the bottom intersection:
- ∠4 and ∠5 are vertical angles? No — usually vertical angles are opposite.

Wait — likely:

- ∠4 and ∠5 are adjacent angles forming a linear pair → ∠4 + ∠5 = 180°
- ∠6 is at the top, and if it’s vertically opposite to ∠4? No — different intersections.

Actually, if ∠4 and ∠6 are on the same side of the transversal and same relative position, they are corresponding angles.

But again — we’re given ∠4 ≅ ∠6, and we need to prove ∠5 ≅ ∠6.

Note: ∠4 and ∠5 are vertical angles? Let’s assume from diagram:

At the bottom intersection:
- ∠4 and ∠5 are vertical angles → so ∠4 ≅ ∠5 (always true)

But then if ∠4 ≅ ∠6 (given), then by transitivity, ∠5 ≅ ∠6.

That’s too simple — but possible.

Alternatively, if ∠4 and ∠5 are a linear pair → ∠4 + ∠5 = 180°, and ∠6 is something else.

But given the diagram and common labeling, I think:

> ∠4 and ∠5 are vertical angles → always congruent
> Given: ∠4 ≅ ∠6
> Therefore: ∠5 ≅ ∠6 by transitivity

Let’s go with that.

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Proof:

| Statement | Reason |
|----------|--------|
| 1. ∠4 ≅ ∠6 | Given |
| 2. ∠4 ≅ ∠5 | Vertical Angles Theorem (at bottom intersection) |
| 3. ∠5 ≅ ∠6 | Transitive Property of Congruence (from 1 and 2) |

✔ Conclusion: ∠5 ≅ ∠6

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## ✔ Final Answers Summary:

1. ∠1 ≅ ∠2 — proven using linear pairs and substitution.
2. ∠2 ≅ ∠3 — proven using linear pairs and substitution.
3. ∠2 is supplementary to ∠3 — proven by substituting congruent angles into linear pair equation.
4. ∠5 ≅ ∠6 — proven using vertical angles and transitivity.

All proofs rely on basic angle relationships: linear pairs, vertical angles, substitution, and transitive property.