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| | 13 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Before any theorems will be introduced, try to discover some properties of angles using the interactive applet. While exploring, think about how those properties could be proven. The applet allows for translations and rotations of the angles. Consider a pair of vertical angles.
Vertical angles can be mapped onto each other by using a rotation. Since rotations are rigid motions, the angle measures are preserved. This leads to the following theorem.
Vertical angles are always congruent.
Based on the characteristics of the diagram, the following relations hold true.
∠ 1 ≅ ∠ 3
∠ 2 ≅ ∠ 4
Therefore, by the Angle Addition Postulate, the sum of m∠ 1 and m∠ 2 is 180^(∘), and the sum of m∠ 2 and m∠ 3 is also 180^(∘). These facts can be used to express m∠ 2 in terms of m∠ 1 and in terms of m∠ 3.
| Angle Addition Postulate | Isolate m∠ 2 |
|---|---|
| m∠ 1+m∠ 2 = 180^(∘) | m∠ 2 = 180^(∘)-m∠ 1 |
| m∠ 2+m∠ 3 = 180^(∘) | m∠ 2 = 180^(∘)-m∠ 3 |
By the Transitive Property of Equality, the expressions representing m∠ 2 can be set equal to each other. m∠ 2= 180^(∘)-m∠ 1 m∠ 2= 180^(∘)-m∠ 3 ⇓ 180^(∘)-m∠ 1= 180^(∘)-m∠ 3 Then the equation can be simplified.
LHS-180^(∘)=RHS-180^(∘)
LHS * (- 1)=RHS* (- 1)
By the definition of congruent angles, this means that the vertical angles ∠ 1 and ∠ 3 are congruent angles. The same process can be used to prove ∠ 2 and ∠ 4 congruent.
The previous proof can be summarized in the following two-column table.
Statements
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Reasons
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1. 1. l_1 and l_2 lines
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1. 1. Given
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2. 2. ∠ 1 and ∠ 2 supplementary
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2. 2. Definition of straight angle
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3. 3. m∠ 1+m∠ 2=180^(∘)
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3. 3. Definition of supplementary angles
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4. 4. m∠ 2=180^(∘)-m∠ 1
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4. 4. Subtraction Property of Equality
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5. 5. ∠ 2 and ∠ 3 supplementary
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5. 5. Definition of straight angle
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6. 6. m∠ 2+m∠ 3=180^(∘)
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6. 6. Definition of supplementary angles
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7. 7. m∠ 2=180^(∘)-m∠ 3
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7. 7. Subtraction Property of Equality
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8. 8. 180^(∘)-m∠ 1=180^(∘)-m∠ 3
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8. 8. Transitive Property of Equality
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9. 9. m∠ 1=m∠ 3
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9. 9. Subtraction and Multiplication Properties of Equality
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Suppose that points A and B are rotated 180^(∘) about point E.
The points A and B are mapped onto the points A' and B' after the rotation. This means that ∠ AEB is mapped onto ∠ A'EB'. Since rotations are a rigid motion, ∠ AEB and ∠ A'EB' are congruent angles. ∠ AEB ≅ ∠ A'EB' Since the point A' lies on EC and point B' lies on ED, ∠ A'EB' is congruent to ∠ CED. ∠ A'EB' ≅ ∠ CED By applying the Transitive Property of Congruence, it can be confirmed that ∠ AEB is congruent to ∠ CED. ∠ AEB ≅ ∠ A'EB' ∠ A'EB' ≅ ∠ CED ⇓ ∠ AEB ≅ ∠ CED
In Flowerland Village, there is a crossroad between Tulip Street and Rose Street. There is a plan to continue the construction of Tulip Street toward the southwest. At the moment, the crossroad forms two angles, whose measures are expressed by 3x+14 and 6x-5, respectively.
Now that the value of x is known, the measures of ∠ 1 and ∠ 2 can be calculated.
| x=19 | |
|---|---|
| m∠ 1=3x+14 | m∠ 2=6x-5 |
| m∠ 1=3( 19)+14 | m∠ 2=6( 19)-5 |
| m∠ 1=57+14 | m∠ 2=114-5 |
| m∠ 1= 71^(∘) | m∠ 2= 109^(∘) |
Next, by analyzing the position of ∠ 1 and ∠ 3, as well as ∠ 2 and ∠ 4, it can be noted that these are vertical angles. Therefore, by the Vertical Angles Theorem, they are two pairs of congruent angles. lm∠ 1=m∠ 3 m∠ 2=m∠ 4 ⇒ lm∠ 3= 71^(∘) m∠ 4= 109^(∘) In this way, it was obtained that ∠ 1 and ∠ 3 are each 71^(∘), and ∠ 2 and ∠ 4 are each 109^(∘).
Consider two parallel lines cut by a transversal. The applet shows a pair of corresponding angles, ∠ A and ∠ B. Is it possible to translate one line so that one of these angles maps onto another?
The observed relation between corresponding angles is presented and proven in the following theorem.
If parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.
If l_1 ∥ l_2, then ∠ 1 ≅ ∠ 5, ∠ 2 ≅ ∠ 6, ∠ 3 ≅ ∠ 7, and ∠ 4 ≅ ∠ 8.
It should be noted that when translating one line onto another, the pairs of corresponding angles overlap and seem to have the same measures. In fact, because a translation is a rigid motion, and a rigid motion preserves length and angle measures, the pairs of corresponding angles are congruent angles.
Note that the converse statement is also true.
If two lines and a transversal form corresponding angles that are congruent, then the lines are parallel.
If ∠ 1 ≅ ∠ 5, ∠ 2 ≅ ∠ 6, ∠ 3 ≅ ∠ 7, or ∠ 4 ≅ ∠ 8, then l_1 ∥ l_2.
Since the goal is to prove that l_1 is parallel to l_2, it will be temporarily assumed that l_1 and l_2 are not parallel. Temporary Assumption l_1 ∦ l_2 By the Parallel Postulate, there exists a line n parallel to l_2 that passes through the point of intersection between l_1 and l_3. This line forms ∠3 and ∠4.
By the Angle Addition Postulate, m∠1 is equal to the sum of m∠3 and m∠4. m∠1=m∠3+m∠4 Since n and l_2 are parallel lines that are cut by a transversal, by the Corresponding Angles Theorem, ∠3 and ∠2 are congruent. By the definition of congruence, these angles have the same measure. ∠3 ≅ ∠2 ⇕ m∠3 = m∠2 By the Substitution Property of Equality, m∠2 can be substituted for m∠3 into the equation for m∠1. m∠1=m∠3+m∠4 ↓ m∠1= m∠2+m∠4 From the above equation and since m∠ 4 is a positive number, it can be concluded that m∠1 is greater than m∠2. m∠1>m∠2 This contradicts the given fact that ∠1 and ∠2 are congruent. The contradiction came from assuming that l_1 and l_2 are not parallel lines. Therefore, l_1 and l_2 must be parallel lines.
In a Flowerland Village house, there are stairs with hand railings like shown in the diagram. The measures of ∠ 1 and ∠ 2 are expressed as 5t-2 and 4t+12, respectively.
What are the measures of ∠ 1 and ∠ 2?
Now that the value of t is known, the measure of each of the angles can be calculated.
Since the angles are congruent, it can be concluded that they both measure to be 68^(∘). l m ∠ 1 = m ∠ 2 m ∠ 1 = 68^(∘) ⇒ m ∠ 2 = 68^(∘)
Like corresponding angles, alternate interior angles are also formed by two parallel lines cut by a transversal.
If parallel lines are cut by a transversal, then alternate interior angles are congruent.
If l_1 ∥ l_2, then ∠ 1 ≅ ∠ 2 and ∠ 3 ≅ ∠ 4.
Notice that by definition ∠ 2 and ∠ 5 are vertical angles. By the Vertical Angles Theorem, they are therefore congruent angles. ∠ 2 ≅ ∠ 5 Furthermore, by definition ∠ 5 and ∠ 1 are corresponding angles. Hence, by the Corresponding Angles Theorem, ∠ 5 and ∠ 1 are also congruent angles. ∠ 5 ≅ ∠ 1 Applying the Transitive Property of Congruence, ∠ 2 and ∠ 1 can be concluded to be congruent angles as well. ∠ 2 ≅ ∠ 5 ∠ 5 ≅ ∠ 1 ⇒ ∠ 2 ≅ ∠ 1 The same reasoning applies to the other pair of alternate interior angles. Therefore, when a pair of parallel lines is cut by a transversal, the pairs of alternate interior angles are congruent.
The previous proof can be summarized in the following two-column table.
Statements
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Reasons
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1. 1. ∠ 2 and ∠ 5 are vertical angles
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1. 1. Def. of vertical angles
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2. 2. ∠ 2≅ ∠ 5
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2. 2. Vertical Angles Theorem
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3. 3. ∠ 5 and ∠ 1 are corresponding angles
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3. 3. Def. of corresponding angles
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4. 4. ∠ 5≅ ∠ 1
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4. 4. Corresponding Angles Theorem
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5. 5. ∠ 2 ≅ ∠ 1
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5. 5. Transitive Property of Congruence
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Next, A, B, C, and D will be translated parallel to the transversal until the points A, C, and D lie on l_2. Then, A, B, and C will be rotated 180^(∘) about F. It should be noted that since point D lies on the transversal, when translating it to l_2 the point will fall into the same position as F. Therefore, D will not be affected by the rotation around F.
After this combination of rigid motions, A, B, C, and D are mapped onto A', B', C', and D'. This means that ∠ ADB is mapped onto ∠ A'D'B'. Therefore, ∠ ADB and ∠ A'D'B' are congruent angles. ∠ ADB ≅ ∠ A'D'B' Note that D' and F share the same location. It can also be seen that A' lies on FG and B' lies on FH. Because of this, ∠ A'D'B' is congruent to ∠ GFH. ∠ A'D'B' ≅ ∠ GFH Applying the Transitive Property of Congruence, ∠ ADB is congruent to GFH. ∠ ADB ≅ ∠ A'D'B' ∠ A'D'B' ≅ ∠ GFH ⇓ ∠ ADB ≅ ∠ GFH
The converse statement is also true.
If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel.
If ∠ 1 ≅ ∠ 2 or ∠ 3 ≅ ∠ 4, then l_1 ∥ l_2.
It needs to be proven that l_1 and l_2 are parallel lines. It is already given that ∠ 1 is congruent to ∠ 2. ∠ 1 ≅ ∠ 2 The diagram shows that ∠ 2 and ∠ α are vertical angles. By the Vertical Angles Theorem, these angles are congruent. ∠ 2 ≅ ∠ α Notice the common angle of ∠ 2 in both relationships. By the Transitive Property of Congruence, since ∠ 1 is congruent to ∠ 2 and ∠ 2 is congruent to ∠ α, then ∠ 1 is congruent ∠ α. ∠ 1 ≅ ∠ 2 ∠ 2 ≅ ∠ α ⇓ ∠ 1 ≅ ∠ α The diagram also shows that ∠ 1 and ∠ α are corresponding angles. Given that relation, the Converse Corresponding Angles Theorem can be applied.
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Converse Corresponding Angles Theorem |
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If two lines are cut by a transversal so that the corresponding angles are congruent, then the lines are parallel. |
Since ∠ 1 and ∠ α are corresponding congruent angles, then l_1 and l_2 are parallel lines. To summarize, all of the steps will be described in a two-column proof.
Statement
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Reason
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1. 1. ∠ 1 ≅ ∠ 2
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1. 1. Given
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2. 2. ∠ 2 ≅ ∠ α
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2. 2. Vertical Angles Theorem
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3. 3. ∠ 1 ≅ ∠ α
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3. 3. Transitive Property of Congruence
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4. 4. l_1 ∥ l_2
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4. 4. Converse Corresponding Angles Theorem
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Similar properties can be discovered for alternate exterior angles.
If parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.
If l_1 ∥ l_2, then ∠ 1 ≅ ∠ 2 and ∠ 3 ≅ ∠ 4.
Notice that by definition, ∠ 2 and ∠ 8 are corresponding angles. Therefore, by the Corresponding Angles Theorem, they are congruent angles. ∠ 2 ≅ ∠ 8 Furthermore, by definition, ∠ 8 and ∠ 1 are vertical angles. Therefore, by the Vertical Angles Theorem, ∠ 8 and ∠ 1 are congruent angles. ∠ 8 ≅ ∠ 1 Then, by applying the Transitive Property of Congruence, ∠ 2 and ∠ 1 can be concluded to be congruent angles as well. ∠ 2 ≅ ∠ 8 ∠ 8 ≅ ∠ 1 ⇒ ∠ 2 ≅ ∠ 1 The same reasoning applies to the other pair of alternate exterior angles. Therefore, when a pair of parallel lines is cut by a transversal, the pairs of alternate exterior angles are congruent.
The previous proof can be summarized in the following two-column table.
Statements
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Reasons
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1. 1. ∠ 2 and ∠ 8 are corresponding angles
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1. 1. Def. of corresponding angles
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2. 2. ∠ 2≅ ∠ 8
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2. 2. Corresponding Angles Theorem
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3. 3. ∠ 8 and ∠ 1 are vertical angles
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3. 3. Def. of vertical angles
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4. 4. ∠ 8≅ ∠ 1
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4. 4. Vertical Angles Theorem
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5. 5. ∠ 2 ≅ ∠ 1
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5. 5. Transitive Property of Congruence
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Next, A, B, C, and D will be translated in the direction of the transversal so that points A, C, and D lie on l_2. Then, A, B, and C will be rotated 180^(∘) about F.
After this combination of rigid motions, A, B, C, and D are mapped onto A', B', C', and D'. This means that ∠ ADB is mapped onto ∠ A'D'B'. Therefore, ∠ ADB and ∠ A'D'B' are congruent angles. ∠ ADB ≅ ∠ A'D'B' Since D' and F share the same location, A' lies on FG and B' lies on FH. Because of this, ∠ A'D'B' is congruent to ∠ GFH. ∠ A'D'B' ≅ ∠ GFH Applying the Transitive Property of Congruence, ∠ ADB is congruent to ∠ GFH. ∠ ADB ≅ ∠ A'D'B' ∠ A'D'B' ≅ ∠ GFH ⇓ ∠ ADB ≅ ∠ GFH It has been proved that one pair of alternate exterior angles is congruent. Further, since C' lies on EF, it can also be proven that ∠ C'D'B' is congruent to ∠ EFH. ∠ C'D'B' ≅ ∠ EFH Applying the Transitive Property of Congruence again, ∠ CDB is congruent to ∠ EFH. ∠ CDB ≅ ∠ C'D'B' ∠ C'D'B' ≅ ∠ EFH ⇓ ∠ CDB ≅ ∠ EFH To conclude, it has been obtained that both pairs of alternate exterior angles are congruent.
∠ ADB ≅ ∠ GFH and ∠ CDB ≅ ∠ EFH
If two lines and a transversal form alternate exterior angles that are congruent, then the two lines are parallel.
If ∠ 1 ≅ ∠ 2 or ∠ 3 ≅ ∠ 4, then l_1 ∥ l_2.
It needs to be proven that l_1 and l_2 are parallel lines. It is already given that ∠ 1 is congruent to ∠ 2. ∠ 1 ≅ ∠ 2 From the diagram, it can also be noted that ∠ 2 and ∠ α are vertical angles. By the Vertical Angles Theorem, these angles are congruent. ∠ 2 ≅ ∠ α By the Transitive Property of Congruence, because ∠ 1 is congruent to ∠ 2 and ∠ 2 is congruent to ∠ α, ∠ 1 is congruent to ∠ α. ∠ 1 ≅ ∠ 2 ∠ 2 ≅ ∠ α ⇓ ∠ 1 ≅ ∠ α Further, ∠ 1 and ∠ α are corresponding angles. Hence, the Converse Corresponding Angles Theorem can be applied.
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Converse Corresponding Angles Theorem |
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If two lines are cut by a transversal so that the corresponding angles are congruent, then the lines are parallel. |
Since ∠ 1 and ∠ α are corresponding congruent angles, l_1 and l_2 are parallel lines. Each step of the proof will now be summarized in a two-column proof.
Statement
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Reason
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1. 1. ∠ 1 ≅ ∠ 2
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1. 1. Given
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2. 2. ∠ 2 ≅ ∠ α
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2. 2. Vertical Angles Theorem
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3. 3. ∠ 1 ≅ ∠ α
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3. 3. Transitive Property of Congruence
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4. 4. l_1 ∥ l_2
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4. 4. Converse Corresponding Angles Theorem
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In order to build Tulip Street on the south side of the Lilian river, which goes through Flowerland Village, there is a need to build a bridge. Devontay, an architect, proposed the following plan for the bridge.
It is known that the measure of ∠ 1 is equal to 4a+11 and the measure of ∠ 2 is equal to 8a-53. What are the measures of ∠ 1 and ∠ 2?
Therefore, by the Alternate Interior Angles Theorem, these angles are congruent. Hence, the measures of these angles are the same. ∠ 1&≅ ∠ 2 &⇓ m∠ 1&=m∠ 2 By substituting m∠ 1 with 4a+11 and m∠ 2 with 8a-53, the equation for a can be formed.
m∠ 1= 4a+11, m∠ 2= 8a-53
LHS-4a=RHS-4a
LHS+53=RHS+53
.LHS /4.=.RHS /4.
Rearrange equation
Knowing the value of a, the measure of each of these angles can be calculated.
Since the angles are congruent, it can be concluded that they both measure 75^(∘). l m∠ 1= m ∠ 2 m ∠ 1 = 75 ^(∘) ⇒ m ∠ 2 = 75 ^(∘)
Up to now, some basic theorems about angles have been seen and proven through some rigid motions. Before the end of the lesson, one last theorem about segments will be learned. Consider a perpendicular bisector of a segment AB. Move C and the endpoints of the segment and compare the distances AC and BC.
Any point on a perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.
Based on the characteristics of the diagram, CM is the perpendicular bisector of AB. Therefore, C is equidistant from A and B.
AC=BC
Consider a triangle with vertices A, M, and C, and another triangle with vertices and B, M, and C.
Both △ ACM and △ BCM have a right angle and congruent legs AM and BM. Since all right angles are congruent, ∠ AMC≅ ∠ BMC. Furthermore, by the Reflexive Property of Congruence, CM is congruent to itself.
By the Side-Angle-Side Congruence Theorem, the triangles are congruent. Therefore, since corresponding parts of congruent figures are congruent, their hypotenuses AC and BC are also congruent. By the definition of congruent segments, AC and BC have the same length. This means that C is equidistant from A and B.
Using this reasoning it can be proven that any point on a perpendicular bisector is equidistant from the endpoints of the segment.
The proof can be summarized in the following two-column table.
Statements
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Reasons
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1. 1. AM ≅ MB ∠ AMC and ∠ BMC are right angles |
1. 1. Definition of a perpendicular bisector.
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2. 2. ∠AMC≅ ∠BMC
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2. 2. All right angles are congruent.
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3. 3. CM≅ CM
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3. 3. Reflexive Property of Congruence.
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4. 4. △ ACM ≅ △ BCM
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4. 4. SAS Congruence Theorem.
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5. 5. AC ≅ BC
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5. 5. Corresponding parts of congruent figures are congruent.
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6. 6. AC = BC
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6. 6. Definition of congruent segments.
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Using the given points A, B, C, and M as vertices, two triangles can be formed. The resulting triangles, △ ACM and △ BCM, can be proven to be congruent by identifying a congruence transformation that maps one triangle onto the other.
Since AM and BM are congruent, the distance between A and M is equal to the distance between B and M. Therefore, A is the image of B after a reflection across CM.
Since C lies on CM, a reflection across CM maps C onto itself. The same is true for M.
| Reflection Across CM | |
|---|---|
| Preimage | Image |
| B | A |
| C | C |
| M | M |
The above table shows that the images of the vertices of △ BCM are the vertices of △ ACM. Therefore, △ ACM is the image of △ BCM after a reflection across CM. Since a reflection is a rigid motion, this proves that the triangles are congruent.
Corresponding parts of congruent figures are congruent, so AC and BC are congruent. By the definition of congruent segments, AC and BC are equal. This means that C is equidistant from A and B. AC=BC The same reasoning can be applied to any point on a perpendicular bisector, showing that the point is equidistant from the endpoints of the segment.
If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment.
AC=CB ⇓ CM⊥ AB and AM=MB
To prove that C lies on the perpendicular bisector of AB, it will be shown that the line perpendicular to AB through C bisects AB. If M is the point of intersection between the line and the segment, it must be proven that AM=MB.
This line forms two right triangles that share a common leg CM. Because all right angles are congruent, ∠ AMC is congruent to ∠ BMC. Also, by the Reflexive Property of Congruence, CM is congruent to itself. Since AC is equal to BC, AC is congruent to BC. ∠ AMC ≅ ∠ BMC CM≅CM AC≅BC By the Hypotenuse Leg Theorem, △ AMC and △ BMC are congruent triangles. Because corresponding parts of congruent figures are congruent, AM is congruent to BM.
Additionally, it was already known that CM and AB are perpendicular.
CM⊥ AB and AM=MB
By the definition of a perpendicular bisector, CM is the perpendicular bisector of AB. Therefore, C lies on the perpendicular bisector of AB.
In Flowerland Village, there are two related families, Funnystongs and Cleverstongs, who live opposite each other. Mr. Funnystong and Mr. Cleverstong want to pave a road between the houses so that every point of the road is equidistant to their houses.
If the houses are 16 meters away, how far from the houses and along what line should the road be paved?
Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.
With this theorem in mind, the position of the road can be determined. To do so, draw a segment whose endpoints are located at the houses.
Before drawing the perpendicular bisector of this segment, its midpoint should be found. Since the distance between the houses is 16 meters, the perpendicular bisector will pass through a point that is 162=8 meters away from the houses.
Based on the theorem, it can be said that each point on the bisector is equidistant from the houses. Therefore, the road between the houses should be paved along the segment's perpendicular bisector.
Determine the value of θ.
Examining the diagram, we see that the angle that measures θ and ∠ ABC form a pair of corresponding angles. Since l ∥ k, we know that these angles are congruent by the Corresponding Angles Theorem.
Notice that ∠ ABC and the angle that measures (x+10^(∘)) form a linear pair. Since a linear pair are supplementary angles, we can write the following equation. θ+(x+10^(∘) )=180^(∘) ⇓ θ+x=170^(∘) Since this equation contains two variables, we need a second equation that contains these variables in order to find their values. Note that the angle that measures x also forms a linear pair with ∠ ACB. Therefore, we know that ∠ ACB equals 180^(∘)-x.
Using the Interior Angles Theorem, we can write a second equation that contains both θ and x. θ+70^(∘)+(180^(∘)-x)=180^(∘) ⇓ θ-x=- 70^(∘) Now we have two equations that both contain θ and x. If we combine them, we get a system of equations. θ+x=170^(∘) θ-x=- 70^(∘) Let's solve this system by using the Elimination Method.
We found that the value of θ is 50^(∘). There is no need to solve the first equation.
Kriz bent a metal wire into the following shape.
What is the measure of ∠ BCD?
The fastest way of solving this exercise is to draw a third segment through C that is parallel to both ED and BA.
From here we can identify two pairs of alternate interior angles, ∠ EDC and ∠ GCD, and ∠ ABC and ∠ FCB. Since ED∥ FG and BA∥ FG, we know that these alternate interior angles are congruent.
Notice that ∠ DCF and ∠ GCD form a straight angle. With this information, we can write an equation and solve it for v.
Now that we know the value of v, we can determine the measure of ∠ BCD. m∠ BCD=8( 10^(∘))=80^(∘)
To find the measure of ∠ BCD, we can also extend a few segments, creating a triangle.
Here we can identify three different angle pairs. Angles:& ∠ EDC and ∠ CDF [-0.2em] Relationship:& Linear Pair [1em] Angles:& ∠ FCD and ∠ BCD [-0.2em] Relationship:& Linear Pair [1em] Angles:& ∠ ABC and ∠ DFC [-0.2em] Relationship:& Alternate interior angles Let's go through each type of angle pair.
Let's add this information to the diagram.
Now we can use the Interior Angles Theorem to write an equation. 50^(∘)+3v+( 180^(∘)-8v)=180^(∘) Let's solve this equation for v.
Now we can calculate the measure of ∠ BCD. m∠ BCD=8(10^(∘))=80^(∘)