module 2 mid topic 2 assessment review
period: 1-2
date: jun/7
1.) identify the property that justifies each statement below.
(a) if m∠1 = 134° and m∠2 = 134°, then m∠1 = m∠2
(b) if gh = jk, then gh - rs = jk - rs
(c) if ∠efg ≅ ∠lmn and ∠lmn ≅ ∠spt, then ∠efg ≅ ∠spt
(d) m∠pqr = m∠pqr
2.) the figure to the right shows lines that intersect at point o.
a. name a pair of vertical angles.
b. name a pair of supplementary angles.
(image of intersecting lines at o with points e, p on one line, a, b, c, d on others)

Question

module 2 mid topic 2 assessment review
period: 1-2
date: jun/7
1.) identify the property that justifies each statement below.
(a) if m∠1 = 134° and m∠2 = 134°, then m∠1 = m∠2
(b) if gh = jk, then gh - rs = jk - rs
(c) if ∠efg ≅ ∠lmn and ∠lmn ≅ ∠spt, then ∠efg ≅ ∠spt
(d) m∠pqr = m∠pqr
2.) the figure to the right shows lines that intersect at point o.
 a. name a pair of vertical angles.
 b. name a pair of supplementary angles.
(image of intersecting lines at o with points e, p on one line, a, b, c, d on others)

Accepted Answer

### Problem 1(a)
# Explanation:
## Step1: Analyze the statement
We have two angles with the same measure, $ m\angle1 = 134^\circ $ and $ m\angle2 = 134^\circ $, and we conclude $ m\angle1 = m\angle2 $. This is the **Substitution Property** (or more specifically, the property of equality that if two quantities are equal to the same quantity, they are equal to each other, which is a form of the Substitution or Transitive? Wait, no, here both are equal to $ 134^\circ $, so it's the **Substitution Property** or the **Transitive Property of Equality**? Wait, the Transitive Property is if $ a = b $ and $ b = c $, then $ a = c $. Here, $ m\angle1 = 134^\circ $ and $ m\angle2 = 134^\circ $, so $ m\angle1 = m\angle2 $ by the **Substitution Property** (replacing $ 134^\circ $ in the first equation with $ m\angle2 $ from the second). Alternatively, it's the **Reflexive? No, Reflexive is $ a = a $. Wait, no, this is the **Transitive Property of Equality**? Wait, no, Transitive is $ a = b $, $ b = c $, so $ a = c $. Here, $ m\angle1 = 134^\circ $ and $ m\angle2 = 134^\circ $, so $ m\angle1 = m\angle2 $ because they are both equal to the same number. This is the **Substitution Property** or the **Property of Equality (Transitive)**? Wait, actually, the correct property here is the **Transitive Property of Equality**? No, Transitive is for three terms. Wait, the correct property is the **Substitution Property** or the **Property that if two quantities are equal to the same quantity, they are equal to each other**, which is a form of the **Transitive Property**? Wait, maybe I'm overcomplicating. The statement is: If $ m\angle1 = 134^\circ $ and $ m\angle2 = 134^\circ $, then $ m\angle1 = m\angle2 $. This is the **Transitive Property of Equality**? No, Transitive is $ a = b $, $ b = c $, so $ a = c $. Here, $ a = m\angle1 $, $ b = 134^\circ $, $ c = m\angle2 $. So $ a = b $, $ c = b $, so $ a = c $ by Transitive. Yes, that's the Transitive Property of Equality. Wait, but sometimes this is called the **Substitution Property**. Alternatively, it's the **Property of Equality (Transitive)**.

Wait, let's recall:

- **Reflexive Property**: $ a = a $
- **Symmetric Property**: If $ a = b $, then $ b = a $
- **Transitive Property**: If $ a = b $ and $ b = c $, then $ a = c $
- **Substitution Property**: If $ a = b $, then $ a $ can be replaced with $ b $ in any equation.

In this case, $ m\angle1 = 134^\circ $ and $ m\angle2 = 134^\circ $, so we can substitute $ 134^\circ $ in the first equation with $ m\angle2 $ (from the second equation), giving $ m\angle1 = m\angle2 $. So this is the **Substitution Property**.

## Step2: Confirm the property
The statement uses the fact that both angles are equal to $ 134^\circ $, so we can substitute one for the other, leading to $ m\angle1 = m\angle2 $. This is justified by the **Substitution Property of Equality** (or the Transitive Property, but more accurately, Substitution here).

# Answer:
Substitution Property (or Transitive Property of Equality, but more accurately, Substitution)

### Problem 1(b)
# Explanation:
## Step1: Analyze the statement
We have $ GH = JK $, and we conclude $ GH - RS = JK - RS $. This is the **Subtraction Property of Equality**, which states that if $ a = b $, then $ a - c = b - c $ for any real number $ c $.

## Step2: Apply the property
Here, $ a = GH $, $ b = JK $, and $ c = RS $. Since $ GH = JK $, subtracting $ RS $ from both sides gives $ GH - RS = JK - RS $, which is the Subtraction Property of Equality.

# Answer:
Subtraction Property of Equality

### Problem 1(c)
# Explanation:
## Step1: Analyze the statement
We have $ \angle EFG \cong \angle LMN $ and $ \angle LMN \cong \angle SPT $, and we conclude $ \angle EFG \cong \angle SPT $. This is the **Transitive Property of Congruence**, which states that if $ \angle A \cong \angle B $ and $ \angle B \cong \angle C $, then $ \angle A \cong \angle C $.

## Step2: Apply the property
Here, $ \angle EFG \cong \angle LMN $ (first congruence) and $ \angle LMN \cong \angle SPT $ (second congruence), so by the Transitive Property of Congruence, $ \angle EFG \cong \angle SPT $.

# Answer:
Transitive Property of Congruence

### Problem 1(d)
# Explanation:
## Step1: Analyze the statement
We have $ m\angle PQR = m\angle PQR $. This is the **Reflexive Property of Equality**, which states that any quantity is equal to itself, i.e., $ a = a $ for any real number $ a $.

## Step2: Apply the property
Here, $ a = m\angle PQR $, so $ m\angle PQR = m\angle PQR $ by the Reflexive Property of Equality.

# Answer:
Reflexive Property of Equality

### Problem 2(a)
# Explanation:
## Step1: Recall vertical angles
Vertical angles are opposite angles formed by the intersection of two lines. They are congruent and have equal measures. In the figure, lines intersect at $ O $, so angles like $ \angle EOB $ and $ \angle PO D $? Wait, looking at the figure: points $ E $, $ O $, $ P $ are colinear (vertical line), and points $ C $, $ O $, $ D $ and $ A $, $ O $, $ B $ are colinear? Wait, the figure shows lines intersecting at $ O $: $ EO P $ is a vertical line, $ C O D $ and $ A O B $ are other lines. So vertical angles are opposite each other. For example, $ \angle EOC $ and $ \angle POD $? Wait, no, let's see: when two lines intersect, vertical angles are the pairs opposite each other. So if we have lines $ AB $ and $ CD $ intersecting at $ O $, and line $ EP $ intersecting them, but the main vertical angles (from the intersection of two lines) would be, for example, $ \angle EOA $ and $ \angle POB $? Wait, maybe a better pair: $ \angle EOB $ and $ \angle POA $? No, let's look at the labels: $ E $ is top, $ P $ is bottom (vertical line), $ C $ is left-top, $ D $ is right-bottom, $ A $ is left-bottom, $ B $ is right-top. So lines $ CA $ and $ DB $ intersect at $ O $, and line $ EP $ intersects at $ O $. So vertical angles from the intersection of $ CA $ and $ DB $ would be $ \angle COE $ and $ \angle DOP $? No, maybe $ \angle AOC $ and $ \angle BOD $? Wait, actually, the standard vertical angles are formed by two intersecting lines. Let's take lines $ AB $ (with points $ A $ and $ B $) and $ CD $ (with points $ C $ and $ D $) intersecting at $ O $. Then vertical angles are $ \angle AOC $ and $ \angle BOD $, or $ \angle AOD $ and $ \angle BOC $. But in the figure, $ E $ and $ P $ are on a vertical line, so maybe $ \angle EOB $ and $ \angle POA $ are vertical angles? Wait, no, let's think again. Vertical angles are opposite each other when two lines cross. So if we have line $ EP $ (vertical) and line $ AB $ (slanted from $ A $ to $ B $), then the vertical angles would be $ \angle EOB $ and $ \angle POA $, or $ \angle EOA $ and $ \angle POB $. Alternatively, from the intersection of $ CD $ (from $ C $ to $ D $) and $ EP $, vertical angles would be $ \angle EOC $ and $ \angle POD $, or $ \angle EOD $ and $ \angle POC $. But a common pair is $ \angle AOC $ and $ \angle BOD $ (from the intersection of $ AB $ and $ CD $). Wait, maybe the answer is $ \angle AOC $ and $ \angle BOD $, or $ \angle EOA $ and $ \angle POB $. Let's pick a simple pair: $ \angle AOC $ and $ \angle BOD $ (vertical angles), or $ \angle EOB $ and $ \angle POA $. Alternatively, $ \angle EOC $ and $ \angle POD $. But the most straightforward pair is $ \angle AOC $ and $ \angle BOD $, or $ \angle EOA $ and $ \angle POB $. Let's go with $ \angle AOC $ and $ \angle BOD $ (but maybe the labels are $ \angle COE $ and $ \angle DOP $, but I think a correct pair is $ \angle AOC $ and $ \angle BOD $).

## Step2: Confirm vertical angles
Vertical angles are formed by two intersecting lines and are opposite each other, with equal measures. So a valid pair is $ \angle AOC $ and $ \angle BOD $ (or any other pair of opposite angles formed by intersecting lines).

# Answer:
$ \angle AOC $ and $ \angle BOD $ (or another valid pair like $ \angle EOB $ and $ \angle POA $)

### Problem 2(b)
# Explanation:
## Step1: Recall supplementary angles
Supplementary angles are two angles whose measures add up to $ 180^\circ $. They can be adjacent (forming a linear pair) or non-adjacent. In the figure, adjacent angles forming a linear pair are supplementary. For example, $ \angle EOB $ and $ \angle BOP $ (since $ EO $ and $ PO $ are a straight line, so they form a linear pair, hence supplementary). Or $ \angle AOC $ and $ \angle COE $, since they form a linear pair along line $ EP $. Wait, line $ EP $ is straight, so any two adjacent angles on $ EP $ are supplementary. For example, $ \angle EOC $ and $ \angle COA $ (since $ EO $ to $ CO $ to $ AO $ is a straight line? Wait, $ E $ is top, $ P $ is bottom, $ C $ is left-top, $ A $ is left-bottom. So $ E $---$ C $---$ A $---$ P $ is a straight line? No, $ E $ is top, $ P $ is bottom (vertical line), $ C $ is on a line from $ C $ to $ D $, $ A $ is on a line from $ A $ to $ B $. Wait, maybe $ \angle EOB $ and $ \angle BOP $ are supplementary because they form a linear pair (since $ EO $ and $ PO $ are a straight line, so $ \angle EOB + \angle BOP = 180^\circ $).

## Step2: Identify a pair
A valid pair of supplementary angles is $ \angle EOB $ and $ \angle BOP $ (since they form a linear pair, hence supplementary) or $ \angle AOC $ and $ \angle COE $, etc.

# Answer:
$ \angle EOB $ and $ \angle BOP $ (or another valid pair like $ \angle AOC $ and $ \angle COE $)

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QUESTION IMAGE

Question

Explanation:

Problem 1(a)

Step1: Analyze the statement

Step2: Confirm the property

Step1: Analyze the statement

Step2: Apply the property

Step1: Analyze the statement

Step2: Apply the property

Answer:

Problem 1(b)