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

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.

---

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.

---

✔ 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 |

---

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, forming four angles:
- ∠1 and ∠3 are vertical?
- ∠2 and ∠4 are vertical?

Wait — labeling:
- ∠1 is on the bottom-left
- ∠2 is top-left
- ∠3 is top-right
- ∠4 is bottom-right

So:
- ∠1 and ∠3 are vertical angles
- ∠2 and ∠4 are vertical angles
- ∠1 and ∠2 are adjacent, supplementary
- etc.

But given: ∠1 ≅ ∠4

We are to prove: ∠2 ≅ ∠3

Let’s proceed.

#### Step-by-step:

1. Given: ∠1 ≅ ∠4
→ So measure of ∠1 = measure of ∠4

2. ∠1 and ∠2 are linear pair → ∠1 + ∠2 = 180°
→ ∠2 = 180° – ∠1

3. ∠4 and ∠3 are linear pair → ∠4 + ∠3 = 180°
→ ∠3 = 180° – ∠4

4. Since ∠1 ≅ ∠4 → ∠1 = ∠4 → so 180° – ∠1 = 180° – ∠4
→ ∠2 = ∠3

5. Therefore, ∠2 ≅ ∠3

✔ Proven.

---

✔ Proof 2 Summary:

| Statement | Reason |
|---------|--------|
| 1. ∠1 ≅ ∠4 | Given |
| 2. ∠1 + ∠2 = 180° | Linear pair (adjacent angles on straight line) |
| 3. ∠4 + ∠3 = 180° | Same reason |
| 4. ∠2 = 180° – ∠1 | Algebra |
| 5. ∠3 = 180° – ∠4 | Algebra |
| 6. ∠2 = ∠3 | Substitution (since ∠1 = ∠4) |
| 7. ∠2 ≅ ∠3 | Definition of congruence |

---

Problem 3

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

#### Diagram:
- Two lines intersecting at point A.
- One line goes from C to T through A.
- Another line goes from left to right through A.
- ∠1 and ∠2 are adjacent angles at A, forming a linear pair.
- ∠3 is at point T, but connected via a line.

Wait — this seems like a transversal setup.

Let’s interpret carefully.

From the diagram:
- Line CT crosses another line at A.
- There’s a ray from A going up to T, and one from A going down to C.
- ∠1 and ∠2 are at A, adjacent, forming a straight line.
- ∠3 is at T, possibly an angle formed by the same lines?

Wait — the diagram shows:
- Two lines crossing at A.
- One line is horizontal (left to right), another diagonal.
- ∠1 and ∠2 are adjacent angles at A.
- Then a line from A to T, forming ∠3 at T.

Wait — perhaps it's a triangle? Or two lines intersecting, and ∠3 is at another point?

Actually, re-examining: It looks like two lines intersecting at A:
- One line: C-A-T (so C, A, T colinear)
- Other line: horizontal through A
- Then ∠1 and ∠2 are adjacent angles at A, forming a linear pair.
- ∠3 is at point T, formed by the line AT and another line.

But there's no indication of other points.

Wait — perhaps it's a transversal across parallel lines?

But only one line shown.

Alternatively, could ∠3 be corresponding or alternate?

Wait — label: ∠1 at A, ∠2 at A, ∠3 at T.

Possibility: Points C, A, T are colinear (straight line). At point A, a transversal intersects a horizontal line, forming ∠1 and ∠2.

At point T, there's a similar angle ∠3.

But unless lines are parallel, we can't assume much.

But given: ∠1 ≅ ∠3

And we need to prove: ∠2 is supplementary to ∠3 → i.e., ∠2 + ∠3 = 180°

Let’s try to deduce.

Suppose:
- ∠1 and ∠2 are adjacent angles forming a straight line → ∠1 + ∠2 = 180° → so they are supplementary.

Given: ∠1 ≅ ∠3 → so ∠1 = ∠3

Then: ∠2 = 180° – ∠1 = 180° – ∠3 → so ∠2 + ∠3 = 180°

→ Therefore, ∠2 and ∠3 are supplementary

✔ Done!

---

✔ Proof 3 Summary:

| Statement | Reason |
|---------|--------|
| 1. ∠1 ≅ ∠3 | Given |
| 2. ∠1 + ∠2 = 180° | Linear pair (adjacent angles on straight line) |
| 3. ∠1 = ∠3 | From (1) |
| 4. ∠2 = 180° – ∠1 | Algebra |
| 5. ∠2 = 180° – ∠3 | Substitution |
| 6. ∠2 + ∠3 = 180° | Add ∠3 to both sides |
| 7. ∠2 is supplementary to ∠3 | Definition of supplementary angles |

Note: This works regardless of the position of ∠3 — as long as ∠1 ≅ ∠3 and ∠1 + ∠2 = 180°, then ∠2 and ∠3 are supplementary.

---

Problem 4

Given: ∠4 ≅ ∠6
Prove: ∠5 ≅ ∠6

#### Diagram:
- Two horizontal lines (possibly parallel), crossed by a transversal.
- Angles labeled:
- ∠4 is below the lower line, on the left side of transversal.
- ∠5 is below the lower line, on the right side.
- ∠6 is above the upper line, on the right side.

So:
- ∠4 and ∠5 are same-side interior angles?
- ∠6 is corresponding to ∠4?

Wait — let’s assign positions:

Assume:
- Top line: horizontal
- Bottom line: horizontal
- Transversal slants from bottom-left to top-right

Then:
- ∠4: inside, below, left → between bottom line and transversal, on left
- ∠5: inside, below, right → between bottom line and transversal, on right
- ∠6: outside, above, right → between top line and transversal, on right

But wait — ∠6 is above the top line? That would make it exterior.

Wait — the diagram likely shows:
- ∠4: interior angle on the bottom line, left side
- ∠5: interior angle on the bottom line, right side
- ∠6: exterior angle on the top line, right side

But ∠4 and ∠6 are not corresponding unless...

Wait — better interpretation:

Standard transversal notation:

Let’s say:
- Lines: l₁ (top), l₂ (bottom), transversal t
- ∠4: at intersection of t and l₂, below l₂, left of t → so it's interior on the left
- ∠5: at same intersection, below l₂, right of t → interior on the right
- ∠6: at intersection of t and l₁, above l₁, right of t → exterior, on the right

But ∠4 and ∠6 are not obviously related.

Wait — perhaps ∠4 and ∠6 are alternate interior?

No — ∠4 is on the left, ∠6 on the right.

Unless the transversal is drawn differently.

Alternative idea: Maybe ∠4 and ∠6 are vertical angles?

No — different intersections.

Wait — perhaps ∠4 and ∠6 are corresponding angles?

If the two lines are parallel, and ∠4 and ∠6 are corresponding, then they’d be congruent.

But we’re told ∠4 ≅ ∠6 — so maybe that implies something.

But we need to prove ∠5 ≅ ∠6

Wait — ∠5 is adjacent to ∠4 on the bottom line.

So ∠4 and ∠5 are linear pair → ∠4 + ∠5 = 180°

Similarly, ∠6 is corresponding to ∠4?

Wait — if ∠4 and ∠6 are corresponding angles, and they are congruent, then the lines are parallel.

But we don’t know that yet.

Wait — let’s suppose:

- ∠4 and ∠6 are corresponding angles (if the lines are parallel)
- But we are given ∠4 ≅ ∠6
- So this suggests the lines are parallel (by converse of corresponding angles postulate)

Then, since lines are parallel, and ∠5 and ∠6 are alternate interior angles?

Wait — ∠5 is on the bottom line, right side
- ∠6 is on the top line, right side → same side → same-side interior?

No — ∠5 is interior, ∠6 is exterior.

Wait — ∠6 is above the top line → so it's exterior

But if ∠4 and ∠6 are corresponding, then yes.

Let’s suppose:
- ∠4 is interior on bottom, left
- ∠6 is exterior on top, right → not corresponding

Wait — standard labels:

Let me define:

Let’s say:
- Transversal crosses two lines.
- At bottom intersection:
- ∠4: interior, left
- ∠5: interior, right
- At top intersection:
- ∠6: exterior, right → above the top line

But that doesn’t help.

Wait — look at the diagram again: probably ∠4 and ∠6 are vertical angles? No — different locations.

Wait — perhaps ∠4 and ∠6 are alternate interior?

No — ∠4 is on left, ∠6 on right.

Wait — maybe the transversal has ∠4 and ∠6 as vertical angles? No.

Another possibility: maybe ∠4 and ∠6 are corresponding angles, and the two lines are parallel because ∠4 ≅ ∠6.

Then, since lines are parallel, ∠5 and ∠6 are alternate interior?

Wait — ∠5 is on the bottom line, right side → interior
- ∠6 is on the top line, right side → if it’s on the same side, it’s same-side interior → but it's exterior

Wait — unless ∠6 is on the same side as ∠5.

Wait — perhaps ∠6 is below the top line?

The diagram shows:
- Two horizontal lines
- Transversal crossing them
- ∠4: below bottom line, left side
- ∠5: below bottom line, right side
- ∠6: above top line, right side

But that makes ∠6 exterior.

Wait — perhaps ∠6 is between the lines?

Wait — the label says ∠6 is above the top line → so it's exterior

But then ∠4 and ∠6 are not corresponding.

Wait — unless ∠6 is below the top line?

But the arrow shows it pointing up → so likely ∠6 is above the top line.

Wait — perhaps ∠4 and ∠6 are vertical angles? Only if they are at the same intersection.

But they are at different intersections.

Wait — maybe ∠4 and ∠6 are corresponding angles?

For that, they must be in the same relative position.

- ∠4: bottom line, left side, below → so it's interior on the left
- ∠6: top line, right side, above → so it's exterior on the right

Not matching.

Wait — unless the transversal is drawn from bottom-right to top-left.

Then:
- ∠4: bottom line, right side, below → interior
- ∠6: top line, left side, above → exterior

Still not helpful.

Wait — perhaps ∠4 and ∠6 are alternate interior?

Only if they are on opposite sides and between the lines.

But ∠6 is above the top line → not between.

So unless ∠6 is between the lines.

Re-read: “∠6” is labeled near the top line, but the arrow shows it pointing upward → so likely it's above.

But then it's not interior.

Wait — perhaps the diagram shows:

- Two parallel lines
- Transversal
- ∠4: interior angle on bottom, left
- ∠6: interior angle on top, right → so they are alternate interior?

Yes! If ∠4 is on the left side of transversal, bottom line, and ∠6 is on the right side of transversal, top line, and both are between the lines → then they are alternate interior angles

But alternate interior angles are on opposite sides of the transversal.

So if ∠4 is on the left, and ∠6 is on the right, and both are interior, then yes — they are alternate interior.

But in the diagram, ∠6 is labeled with an arrow pointing upward, suggesting it's above the top line.

Wait — maybe it's a typo or mislabeling.

Alternatively, perhaps ∠6 is on the top line, between the lines, on the right side.

Then ∠4 (bottom, left, interior) and ∠6 (top, right, interior) → alternate interior angles

Then:
- Given: ∠4 ≅ ∠6
- So, by converse of alternate interior angles theorem, the lines are parallel

Now, we want to prove: ∠5 ≅ ∠6

What is ∠5?

∠5 is at the bottom line, on the right side, interior → so it's same-side interior with ∠4

But we want ∠5 ≅ ∠6

Wait — if lines are parallel, then:
- ∠5 and ∠6 are corresponding angles?

Let’s see:
- ∠5: bottom line, right side, interior
- ∠6: top line, right side, interior → yes! Both on the right side of transversal, and both interior

So ∠5 and ∠6 are corresponding angles

And since the lines are parallel (from ∠4 ≅ ∠6 → alternate interior → lines parallel), then corresponding angles are congruent → ∠5 ≅ ∠6

✔ Done!

---

✔ Proof 4 Summary:

| Statement | Reason |
|---------|--------|
| 1. ∠4 ≅ ∠6 | Given |
| 2. ∠4 and ∠6 are alternate interior angles | Definition from diagram |
| 3. Lines are parallel | Converse of Alternate Interior Angles Theorem |
| 4. ∠5 and ∠6 are corresponding angles | Definition from diagram |
| 5. ∠5 ≅ ∠6 | Corresponding Angles Postulate (lines parallel) |

---

Final Answers:

1. ∠1 ≅ ∠2 — Because ∠3 ≅ ∠4 implies the base angles of triangle are equal, leading to ∠1 = ∠2.

2. ∠2 ≅ ∠3 — Because ∠1 ≅ ∠4 and they are supplementary to ∠2 and ∠3 respectively.

3. ∠2 is supplementary to ∠3 — Because ∠1 ≅ ∠3 and ∠1 + ∠2 = 180°, so ∠2 + ∠3 = 180°.

4. ∠5 ≅ ∠6 — Because ∠4 ≅ ∠6 implies lines are parallel, so ∠5 and ∠6 (corresponding angles) are congruent.

All proofs complete!