ready, set, go! name period date ready topic: are you ready for a test on module 5? figure 1 has been rotated 180° about the midpoint in side bc to form figure 2. figure 1 was then translated to the right to form figure 3. 1. use figure 3 to explain how you know the exterior angle ∠bcc is equal to the sum of the 2 remote interior angles ∠bac and ∠abc. 2. use figure 3 to explain how you know the sum of the angles in a triangle is always 180°. 3. use figure 2 to explain how you know the sum of the angles in a quadrilateral is always 360°. 4. use figure 2 to explain how you know that the opposite angles in a parallelogram are congruent. 5. use figure 2 to explain how you know that the opposite sides in a parallelogram are parallel and congruent. 6. use figure 2 to explain how you know that when two parallel lines are crossed by a transversal, the alternate interior angles are congruent. 7. use figure 2 and/or 3 to explain how you know that when two parallel lines are crossed by a transversal, the same - side interior angles are supplementary. need help? visit www.rsgsupport.org mathematics vision project licensed under the creative commons attribution cc by 4.0 mathematicsvisionproject.org 56 mvp mathemati vision proj

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Accepted Answer

### Question 1
# Explanation:
## Step1: Recall Transformations
Figure 1 is rotated $180^\circ$ about midpoint of $BC$ to get Figure 2, then translated right to get Figure 3. So, $\angle BAC \cong \angle A'$ and $\angle ABC \cong \angle A'BC$ (due to rotation and translation preserving angles).
## Step2: Analyze Parallel Lines
In Figure 3, $AB \parallel A'C$ (from translation, so corresponding sides are parallel). Thus, $\angle BAC=\angle A'CC$ (alternate interior angles) and $\angle ABC = \angle A'BC$ (rotation, and $A'B \parallel BC$? Wait, better: The exterior angle $\angle BCC'$ is formed. Since Figure 3 is translation, $AC \parallel A'C'$? Wait, no. Wait, the three angles $\angle BAC$, $\angle ABC$, and $\angle ACB$ sum to $180^\circ$ (triangle angle sum). But $\angle BCC'$ is supplementary to $\angle ACB$ (linear pair), so $\angle BCC' = 180^\circ - \angle ACB$. But also, $\angle BAC + \angle ABC=180^\circ - \angle ACB$ (triangle angle sum). Thus, $\angle BCC'=\angle BAC + \angle ABC$. Alternatively, using the translated triangle: The angle $\angle BAC$ is equal to the angle between $A'C$ and $CC'$ (alternate interior angles, since $AC \parallel A'C$ from translation), and $\angle ABC$ is equal to the angle between $A'B$ and $BC$ (rotation, so $A'B \parallel AC$? Wait, maybe simpler: In Figure 3, the angle $\angle BCC'$ is composed of the angles equal to $\angle BAC$ and $\angle ABC$ due to the translation and rotation, so their sum equals the exterior angle.
# Answer:
By the transformations (rotation and translation) that form Figure 3, the angles $\angle BAC$ and $\angle ABC$ correspond to angles that, when combined, equal $\angle BCC'$. Specifically, since translation preserves parallelism (so $AB \parallel A'C$), $\angle BAC=\angle A'CC$ (alternate interior angles) and $\angle ABC = \angle A'BC$ (rotation). Then, $\angle BCC'=\angle A'BC+\angle A'CC=\angle ABC + \angle BAC$ (by angle addition), showing the exterior angle equals the sum of the two remote interior angles.

### Question 2
# Explanation:
## Step1: Identify Angles on a Line
In Figure 3, the three interior angles of the triangle ($\angle BAC$, $\angle ABC$, $\angle ACB$) lie along the straight line $ACC'$ (since Figure 3 is a translation, the base $AC$ is translated to $A'C'$, so the angles at $C$ form a linear pair? Wait, no. The three angles $\angle BAC$, $\angle ABC$, and $\angle ACB$ are arranged such that when you translate the triangle, the angles $\angle BAC$ and $\angle ABC$ along with $\angle ACB$ form a straight angle (180°). Because the translated triangle's angle $\angle BAC$ and $\angle ABC$ align with the angles adjacent to $\angle ACB$ to form a straight line. So $\angle BAC + \angle ABC + \angle ACB = 180^\circ$, proving the sum of angles in a triangle is $180^\circ$.
# Answer:
In Figure 3, the three interior angles of the triangle ($\angle BAC$, $\angle ABC$, $\angle ACB$) form a straight angle (linear pair) along the line $ACC'$. Since a straight angle measures $180^\circ$, the sum $\angle BAC + \angle ABC + \angle ACB = 180^\circ$, showing the sum of the angles in a triangle is always $180^\circ$.

### Question 3
# Explanation:
## Step1: Figure 2 is a Parallelogram
Figure 2 is formed by rotating Figure 1 $180^\circ$ about the midpoint of $BC$, so $AB \parallel A'C$ and $AC \parallel A'B$ (since rotation by $180^\circ$ makes opposite sides parallel and congruent, forming a parallelogram $ABCA'$).
## Step2: Sum of Angles in Parallelogram
In a parallelogram, consecutive angles are supplementary (e.g., $\angle A + \angle B = 180^\circ$, $\angle B + \angle C = 180^\circ$, etc.). The sum of all four angles: $(\angle A + \angle B)+(\angle C + \angle D)=180^\circ + 180^\circ = 360^\circ$ (since opposite angles in parallelogram are congruent, $\angle A=\angle C$, $\angle B=\angle D$). Alternatively, divide the parallelogram into two triangles (e.g., diagonal $BC$ divides Figure 2 into two congruent triangles, each with angle sum $180^\circ$, so total $180^\circ \times 2 = 360^\circ$).
# Answer:
Figure 2 is a parallelogram (formed by $180^\circ$ rotation, so opposite sides are parallel and congruent). A diagonal (e.g., $BC$) divides the parallelogram into two congruent triangles. Each triangle has an angle sum of $180^\circ$, so the total angle sum for the quadrilateral (parallelogram) is $180^\circ \times 2 = 360^\circ$, showing the sum of the angles in a quadrilateral is always $360^\circ$.

### Question 4
# Explanation:
## Step1: Rotation Forms Parallelogram
Figure 2 is formed by rotating Figure 1 $180^\circ$ about the midpoint of $BC$. So, the vertex $A$ is rotated to $A'$, so $BA = A'C$ and $CA = A'B$ (rotation preserves length). Also, the angle $\angle ABC$ is rotated to $\angle A'CB$, and $\angle BAC$ is rotated to $\angle A'BC$.
## Step2: Prove Opposite Angles Congruent
In the parallelogram $ABCA'$ (from rotation), $\angle ABC$ and $\angle A'CB$? Wait, no. In a parallelogram, opposite angles are congruent. Since rotation by $180^\circ$ maps $\angle A$ to $\angle C$ and $\angle B$ to $\angle A'$? Wait, better: The rotation maps triangle $ABC$ to triangle $A'CB$, so $\angle BAC \cong \angle A'CB$ and $\angle ABC \cong \angle A'BC$. Wait, no, the parallelogram $ABCA'$ has $\angle A = \angle C$ and $\angle B = \angle A'$ (since $AB \parallel A'C$ and $AC \parallel A'B$, so consecutive angles are supplementary, and opposite angles are equal). So, by the rotation, the angle $\angle BAC$ (at $A$) is congruent to the angle at $C$ ( $\angle A'CB$), and $\angle ABC$ (at $B$) is congruent to the angle at $A'$ ( $\angle BA'C$). Thus, opposite angles in the parallelogram are congruent.
# Answer:
Figure 2 is a parallelogram (formed by $180^\circ$ rotation about $BC$’s midpoint, so $AB \parallel A'C$ and $AC \parallel A'B$). In a parallelogram, when two parallel lines ($AB \parallel A'C$) are cut by a transversal ($BC$ or $AC$), alternate interior angles are congruent. Also, the rotation maps $\angle BAC$ to $\angle A'CB$ and $\angle ABC$ to $\angle A'BC$, showing $\angle A \cong \angle C$ and $\angle B \cong \angle A'$ (opposite angles), hence opposite angles in a parallelogram are congruent.

### Question 5
# Explanation:
## Step1: Rotation Preserves Length
Rotating Figure 1 $180^\circ$ about the midpoint of $BC$ means the distance from $A$ to $B$ is equal to the distance from $A'$ to $C$ (rotation preserves length), and the distance from $A$ to $C$ is equal to the distance from $A'$ to $B$.
## Step2: Parallelism from Rotation
The rotation by $180^\circ$ about the midpoint of $BC$ implies that $AB \parallel A'C$ and $AC \parallel A'B$ (since the rotation maps the vector $AB$ to $A'C$, so they are parallel and equal in length). Thus, in the parallelogram $ABCA'$, opposite sides $AB$ and $A'C$ are parallel (from rotation, same direction) and congruent (rotation preserves length), and $AC$ and $A'B$ are parallel and congruent.
# Answer:
Figure 2 is a parallelogram formed by rotating Figure 1 $180^\circ$ about $BC$’s midpoint. Rotation preserves side lengths ($AB = A'C$, $AC = A'B$) and creates parallelism ($AB \parallel A'C$, $AC \parallel A'B$) because a $180^\circ$ rotation maps vectors $AB$ to $A'C$ and $AC$ to $A'B$, making them parallel and congruent. Thus, opposite sides in a parallelogram are parallel and congruent.

### Question 6
# Explanation:
## Step1: Parallel Lines in Figure 2
Figure 2 is a parallelogram, so $AB \parallel A'C$ and $AC \parallel A'B$ (from $180^\circ$ rotation). Let’s take transversal $BC$ cutting $AB \parallel A'C$. Then, the alternate interior angles $\angle ABC$ and $\angle A'CB$ are congruent (since $AB \parallel A'C$ and $BC$ is transversal). Similarly, transversal $AC$ cutting $AC \parallel A'B$ gives alternate interior angles congruent. Thus, when two parallel lines (e.g., $AB \parallel A'C$) are cut by a transversal (e.g., $BC$), alternate interior angles are congruent.
# Answer:
Figure 2 is a parallelogram with $AB \parallel A'C$ (from $180^\circ$ rotation). When transversal $BC$ intersects $AB \parallel A'C$, the alternate interior angles (e.g., $\angle ABC$ and $\angle A'CB$) are congruent (by the rotation, which preserves angle measures and creates parallel lines). This shows that when two parallel lines are cut by a transversal, alternate interior angles are congruent.

### Question 7
# Explanation:
## Step1: Parallel Lines in Figures 2 and 3
Figure 2 is a parallelogram ($AB \parallel A'C$, $AC \parallel A'B$), and Figure 3 is a translation (so $AC \parallel A'C'$). Take transversal $BC$ cutting $AB \parallel A'C$ (Figure 2): consecutive interior angles $\angle ABC$ and $\angle BCA'$ are supplementary (since $AB \parallel A'C$, same - side interior angles sum to $180^\circ$). In Figure 3, translation gives $AC \parallel A'C'$, and transversal $CC'$ cutting them: same - side interior angles sum to $180^\circ$. Also, in the triangle of Figure 3, the same - side interior angles (e.g., $\angle BAC$ and $\angle ABC$ with transversal $AB$?) No, better: In a parallelogram (Figure 2), consecutive angles are supplementary (same - side interior angles), and in translation (Figure 3), parallel lines with transversal have same - side interior angles supplementary. Thus, when two parallel lines are cut by a transversal, same - side interior angles are supplementary.
# Answer:
In Figure 2 (parallelogram, $AB \parallel A'C$), transversal $BC$ creates same - side interior angles ($\angle ABC$ and $\angle BCA'$) that sum to $180^\circ$ (supplementary) because $AB \parallel A'C$. In Figure 3 (translation, $AC \parallel A'C'$), transversal $CC'$ creates same - side interior angles that are supplementary. This shows that when two parallel lines are cut by a transversal, same - side interior angles are supplementary.

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Question

Explanation:

Question 1

Step1: Recall Transformations

Step2: Analyze Parallel Lines

Step1: Identify Angles on a Line

Step1: Figure 2 is a Parallelogram

Step2: Sum of Angles in Parallelogram

Answer:

Question 2