Geometry Proof Worksheet (3) featuring four angle relationship problems with diagrams.

Let's solve each of the four geometry proof problems step by step. We'll use geometric principles such as angle relationships, properties of parallel lines, vertical angles, supplementary angles, and triangle congruence where applicable.

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Problem 1

Given: ∠3 ≅ ∠4
Prove: ∠1 ≅ ∠2

#### Diagram:
- Triangle ABC with point A and B on a straight line extending left and right.
- ∠3 is at vertex A (between triangle and extension).
- ∠4 is at vertex B (between triangle and extension).
- So, AB is a straight line, and AC and BC form a triangle above it.

#### Step-by-step Proof:

1. Given: ∠3 ≅ ∠4
→ These are base angles of triangle ABC.

2. Since ∠3 and ∠4 are base angles of triangle ABC and they are congruent, then by the Isosceles Triangle Theorem, the sides opposite them are equal:
→ AC ≅ BC
→ So, triangle ABC is isosceles with AC = BC.

3. In an isosceles triangle, the base angles are equal. But here, we already know ∠3 ≅ ∠4 — these are not the base angles of the triangle; they're external angles formed between the triangle and the extended line.

Wait — let’s clarify:

Actually, looking at the diagram:
- Point A and B lie on a straight line.
- C is above, forming triangle ABC.
- ∠3 is the angle between segment AC and the extension to the left of A.
- ∠4 is the angle between segment BC and the extension to the right of B.

But since AB is a straight line, the angles at A and B outside the triangle are adjacent to the interior angles of the triangle.

Let’s define:
- Let ∠CAB = ∠A (interior angle at A)
- Let ∠CBA = ∠B (interior angle at B)

Then:
- ∠3 is the exterior angle at A: ∠3 = 180° – ∠A
- ∠4 is the exterior angle at B: ∠4 = 180° – ∠B

Given: ∠3 ≅ ∠4 → so 180° – ∠A = 180° – ∠B → ∠A = ∠B

So, interior angles at A and B are equal → triangle ABC is isosceles with AC = BC.

Now, in triangle ABC, if ∠A = ∠B, then the remaining angles must be related.

But what about ∠1 and ∠2?

Looking at the diagram:
- ∠1 is the interior angle at A (i.e., ∠CAB)
- ∠2 is the interior angle at B (i.e., ∠CBA)

We just showed that ∠A = ∠B → ∠1 ≅ ∠2

✔ Therefore, ∠1 ≅ ∠2 — Q.E.D.

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✔ Proof 1 Summary:

| Statement | Reason |
|---------|--------|
| 1. ∠3 ≅ ∠4 | Given |
| 2. ∠3 = 180° – ∠1, ∠4 = 180° – ∠2 | Linear pair definition |
| 3. 180° – ∠1 = 180° – ∠2 | Substitution from (1) |
| 4. ∠1 = ∠2 | Subtract 180° from both sides |
| 5. ∠1 ≅ ∠2 | Definition of congruent angles |

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Problem 2

Given: ∠1 ≅ ∠4
Prove: ∠2 ≅ ∠3

#### Diagram:
- Two intersecting lines forming an "X" shape.
- Angles labeled: ∠1, ∠2, ∠3, ∠4 around the intersection.
- Likely, ∠1 and ∠4 are vertical angles, or maybe alternate interior?

Wait — actually, the figure shows two lines crossing: one horizontal, one diagonal.

Let’s assume:
- The horizontal line has points going left and right.
- A diagonal line crosses it.
- Angles:
- ∠1: lower-left
- ∠2: upper-left
- ∠3: upper-right
- ∠4: lower-right

Standard labeling: angles around a point.

If ∠1 and ∠4 are vertical angles, but given ∠1 ≅ ∠4, that would make sense only if they are vertical.

But wait — vertical angles are always congruent. So why is it given?

But look: perhaps the figure shows two lines intersecting, and ∠1 and ∠4 are not vertical.

Wait — more likely:
- ∠1 and ∠3 are vertical?
- ∠2 and ∠4 are vertical?

Wait — standard notation: when two lines intersect, they form two pairs of vertical angles.

Assume:
- ∠1 and ∠3 are vertical → ∠1 ≅ ∠3
- ∠2 and ∠4 are vertical → ∠2 ≅ ∠4

But given: ∠1 ≅ ∠4

We need to prove: ∠2 ≅ ∠3

Let’s proceed.

#### Step-by-step:

1. Given: ∠1 ≅ ∠4
2. Vertical angles: ∠1 ≅ ∠3 (since they are vertical)
→ So ∠1 ≅ ∠3
3. From (1) and (2): ∠4 ≅ ∠3 (by transitive property)
4. But ∠2 and ∠4 are vertical → ∠2 ≅ ∠4
5. So ∠2 ≅ ∠3 (transitive again)

Wait — let’s do this carefully.

Let’s label the angles properly:

At the intersection of two lines:
- ∠1 and ∠3 are vertical angles → ∠1 ≅ ∠3
- ∠2 and ∠4 are vertical angles → ∠2 ≅ ∠4

Given: ∠1 ≅ ∠4

From above:
- ∠1 ≅ ∠3 (vertical)
- ∠2 ≅ ∠4 (vertical)

So:
- ∠1 ≅ ∠4 → given
- But ∠1 ≅ ∠3 → so ∠3 ≅ ∠4
- And ∠2 ≅ ∠4 → so ∠2 ≅ ∠3

Thus, ∠2 ≅ ∠3

✔ Proved.

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✔ Proof 2 Summary:

| Statement | Reason |
|---------|--------|
| 1. ∠1 ≅ ∠4 | Given |
| 2. ∠1 ≅ ∠3 | Vertical angles are congruent |
| 3. ∠2 ≅ ∠4 | Vertical angles are congruent |
| 4. ∠3 ≅ ∠4 | Transitive (from 1 and 2) |
| 5. ∠2 ≅ ∠3 | Transitive (from 3 and 4) |

Alternatively, more directly:
From (1), (2), (3):
- ∠1 ≅ ∠4 (given)
- ∠1 ≅ ∠3 → so ∠3 ≅ ∠4
- ∠2 ≅ ∠4 → so ∠2 ≅ ∠3

Yes.

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Problem 3

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

#### Diagram:
- Three rays forming a "zig-zag" path: point C, A, T.
- Lines: CA and AT, with a transversal?
- ∠1 at point A, between CA and some ray
- ∠2 at point A, between the other rays
- ∠3 at point T

Wait — looks like three points: C, A, T.
- Ray from C to A, then from A to T
- Another ray from A going up, forming ∠1 and ∠2 at A
- ∠3 at T

But the key is: ∠1 ≅ ∠3, and we need to prove ∠2 and ∠3 are supplementary (sum to 180°)

Possibility: This could involve parallel lines or transversals.

But no parallel lines indicated.

Wait — perhaps it's a straight line at point A?

Let’s suppose:
- Points C, A, T are colinear? No — because there's a bend at A.

But the figure shows:
- Line from C to A, then from A to T
- At A, two rays: one going to C, one to T, and another ray forming ∠1 and ∠2
- ∠1 and ∠2 are adjacent angles at A
- ∠3 is at T

Wait — maybe it's a triangle?

Wait — better idea: This might be a transversal cutting two lines, but the diagram shows:
- Two lines crossing at A: one from C to T, and another ray from A going upward.
- ∠1 and ∠2 are adjacent angles at A
- ∠3 is at T, possibly corresponding or alternate?

But without clear labels, let's interpret:

Suppose:
- Line CT passes through A
- Another line goes from A to some point above
- ∠1 is between the upward ray and CA
- ∠2 is between the upward ray and AT
- So ∠1 and ∠2 are adjacent angles forming a straight line? Not necessarily.

Wait — look at the arrows:
- Arrow from C to A to T — so C-A-T is a straight line?
- Then a ray from A going upward
- So ∠1 is between CA and the upward ray
- ∠2 is between the upward ray and AT
- So ∠1 and ∠2 are adjacent angles that together form a straight line?

But if C-A-T is a straight line, then ∠1 + ∠2 = 180° → they are supplementary

But we are to prove that ∠2 is supplementary to ∠3, not ∠1.

Given: ∠1 ≅ ∠3

So if ∠1 + ∠2 = 180° (because they form a straight angle), and ∠1 ≅ ∠3, then:

→ ∠3 + ∠2 = 180° → so ∠2 and ∠3 are supplementary

✔ That’s the proof!

So we need to assume that ∠1 and ∠2 are adjacent angles forming a linear pair.

That is, the ray from A splits the straight angle at A into ∠1 and ∠2.

So:

#### Step-by-step:

1. Assume C-A-T is a straight line → so ∠CAT is a straight angle → measure 180°
2. The upward ray divides ∠CAT into ∠1 and ∠2 → so ∠1 + ∠2 = 180°
3. Given: ∠1 ≅ ∠3 → so m∠1 = m∠3
4. Substitute: m∠3 + m∠2 = 180°
5. Therefore, ∠2 and ∠3 are supplementary

✔ Proved.

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✔ Proof 3 Summary:

| Statement | Reason |
|---------|--------|
| 1. C-A-T is a straight line | Assumed from diagram (points collinear) |
| 2. ∠1 + ∠2 = 180° | Linear pair postulate |
| 3. ∠1 ≅ ∠3 | Given |
| 4. m∠1 = m∠3 | Definition of congruent angles |
| 5. m∠3 + m∠2 = 180° | Substitution |
| 6. ∠2 is supplementary to ∠3 | Definition of supplementary angles |

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Problem 4

Given: ∠4 ≅ ∠6
Prove: ∠5 ≅ ∠6

#### Diagram:
- Two horizontal lines (possibly parallel?)
- A transversal cutting them
- Angles labeled:
- ∠4 and ∠5 are on the same side of the transversal, below the lower line
- ∠6 is above the upper line, on the same side of the transversal

Wait — standard labeling:
- Lower line: ∠4 and ∠5 are adjacent angles at the intersection
- Upper line: ∠6 is on the same side

So:
- ∠4 and ∠5 are linear pair → sum to 180°
- ∠6 is on the upper line, corresponding or alternate?

But given: ∠4 ≅ ∠6

We are to prove: ∠5 ≅ ∠6

So:
- ∠4 ≅ ∠6 (given)
- ∠4 + ∠5 = 180° (linear pair)
- So ∠6 + ∠5 = 180° → they are supplementary
- But we want ∠5 ≅ ∠6 → so unless both are 90°, this doesn’t work.

Wait — unless ∠4 and ∠6 are corresponding angles, and the lines are parallel?

But we are not told the lines are parallel.

Wait — perhaps the diagram shows:
- Two horizontal lines
- A transversal
- ∠4 and ∠6 are corresponding angles
- Given ∠4 ≅ ∠6 → implies lines are parallel (if we assume that)
- Then ∠5 and ∠6 are alternate interior or something?

Wait — let’s define:

Let’s say:
- Lower line: angles ∠4 and ∠5 are adjacent, forming a straight line → ∠4 + ∠5 = 180°
- Upper line: ∠6 is on the same side as ∠4
- So ∠4 and ∠6 are corresponding angles

Given: ∠4 ≅ ∠6 → if corresponding angles are congruent, then the lines are parallel

So:
1. ∠4 ≅ ∠6 → corresponding angles are congruent → lines are parallel
2. Now, ∠5 and ∠6: are they alternate interior or same-side interior?

Wait — ∠5 is below the lower line, on the same side as ∠6?

No — ∠5 is on the lower line, and ∠6 is on the upper line, both on the same side of the transversal → so they are same-side interior angles?

But same-side interior angles are supplementary, not necessarily congruent.

But we are to prove ∠5 ≅ ∠6

Only possible if both are 90°, but not guaranteed.

Wait — maybe the diagram shows:
- ∠4 and ∠5 are vertical angles? No — they are adjacent.

Wait — perhaps ∠4 and ∠5 are adjacent on the lower line, so ∠4 + ∠5 = 180°
- ∠6 is on the upper line, and ∠4 ≅ ∠6

So:
- ∠4 ≅ ∠6 → given
- ∠4 + ∠5 = 180° → so ∠6 + ∠5 = 180° → supplementary

But we want ∠5 ≅ ∠6 → so unless both are 90°, impossible.

Unless... wait — maybe ∠5 and ∠6 are vertical angles?

But they’re not at the same vertex.

Wait — perhaps the transversal intersects two lines, and:
- ∠4 and ∠6 are corresponding
- ∠5 and ∠6 are alternate interior?

No — ∠5 is on the lower line, ∠6 on the upper line, on the same side → so they are same-side interior → sum to 180° if lines are parallel.

But we are to prove ∠5 ≅ ∠6

So only possible if ∠5 = ∠6 = 90°

But not given.

Wait — unless the lines are perpendicular?

But nothing says that.

Wait — perhaps I misread.

Wait — look at the diagram:
- ∠4 is on the lower line, below the transversal
- ∠5 is on the lower line, above the transversal → so ∠4 and ∠5 are vertical angles?

No — if they are on the same side of the transversal, they are adjacent.

Wait — standard: when a transversal crosses a line, it forms two pairs of vertical angles.

So at the lower intersection:
- ∠4 and ∠5 are adjacent angles → linear pair → sum to 180°
- At the upper intersection:
- ∠6 is one angle
- Suppose ∠6 is corresponding to ∠4

Given: ∠4 ≅ ∠6 → so corresponding angles congruent → lines are parallel

Then, since lines are parallel, alternate interior angles are congruent.

But which angles are alternate interior?

- ∠5 and ∠6: are they alternate interior?

Let’s see:
- ∠5 is on the lower line, above the transversal
- ∠6 is on the upper line, below the transversal
- So they are on opposite sides of the transversal, and between the lines → yes! Alternate interior angles

So if lines are parallel, then ∠5 ≅ ∠6

But we don't know lines are parallel yet.

But we have: ∠4 ≅ ∠6

And ∠4 and ∠6 are corresponding angles → so if they are congruent, then the lines are parallel (Corresponding Angles Postulate)

Then, since lines are parallel, alternate interior angles are congruent → ∠5 ≅ ∠6

✔ Yes!

So proof:

1. Given: ∠4 ≅ ∠6
2. ∠4 and ∠6 are corresponding angles
3. So, the two lines are parallel (Converse of Corresponding Angles Postulate)
4. ∠5 and ∠6 are alternate interior angles
5. Therefore, ∠5 ≅ ∠6 (Alternate Interior Angles Theorem)

Wait — but ∠5 and ∠6 are on opposite sides of the transversal and between the lines → yes, alternate interior.

But ∠6 is on the upper line, and ∠5 is on the lower line, both on the same side? No — depends.

Wait — if ∠4 is below the transversal on the lower line, and ∠5 is above the transversal on the lower line, then ∠5 and ∠6 are on opposite sides of the transversal.

Yes — so if ∠6 is on the upper line, below the transversal, and ∠5 is on the lower line, above the transversal → then they are alternate interior angles.

Yes.

So:

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✔ Proof 4 Summary:

| Statement | Reason |
|---------|--------|
| 1. ∠4 ≅ ∠6 | Given |
| 2. ∠4 and ∠6 are corresponding angles | Diagram/definition |
| 3. The two lines are parallel | Converse of Corresponding Angles Postulate |
| 4. ∠5 and ∠6 are alternate interior angles | Definition |
| 5. ∠5 ≅ ∠6 | Alternate Interior Angles Theorem |

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✔ Final Answers:

1. Proved: ∠1 ≅ ∠2 using linear pairs and substitution.
2. Proved: ∠2 ≅ ∠3 using vertical angles and transitive property.
3. Proved: ∠2 is supplementary to ∠3 using linear pair and substitution.
4. Proved: ∠5 ≅ ∠6 using parallel lines and alternate interior angles.

All proofs rely on basic geometric postulates and properties.

Let me know if you'd like these written in formal two-column format!