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Apply Theorem 5.9 from Section 5.3, and remember that if A> B then A=B+x for some x> 0.
Statements
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Reasons
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1. PR≅ PQ
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1. Given
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2. ∠PRQ ≅ ∠PQR
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2. Isosceles Triangle Theorem
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3. m∠PRQ = m∠PQR
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3. Definition of Congruent Angles
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4. m∠PRQ = m∠1 + m∠4 and m∠PQR = m∠2 + m∠3
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4. Angle Addition Postulate
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5. m∠1 + m∠4 = m∠2 + m∠3
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5. Substitution
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6. SQ > SR
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6. Given
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7. m∠4 > m∠3
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7. Theorem 5.9
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8. m∠4 = m∠3 + x
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8. Definition of inequality
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9. m∠1 = m∠2 - x
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9. Subtracting equation in 8 from equation in 5
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10. m∠1 + x = m∠2
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10. Solving for m∠2
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11. m∠1 < m∠2
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11. Definition of inequality
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Let's begin by highlighting the given information in the diagram.
By applying the Isosceles Triangle Theorem we obtain that ∠PRQ ≅ ∠PQR, which implies that m∠PRQ = m∠PQR. Also, by the Angle Addition Postulate we can write the following two equations.
m∠PRQ = m ∠1 + m ∠4
m∠PQR = m ∠2 + m ∠3
In the table below, we will summarize the proof we just did before.
Statements
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Reasons
|
1. PR≅ PQ
|
1. Given
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2. ∠PRQ ≅ ∠PQR
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2. Isosceles Triangle Theorem
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3. m∠PRQ = m∠PQR
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3. Definition of Congruent Angles
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4. m∠PRQ = m∠1 + m∠4 and m∠PQR = m∠2 + m∠3
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4. Angle Addition Postulate
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5. m∠1 + m∠4 = m∠2 + m∠3
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5. Substitution
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6. SQ > SR
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6. Given
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7. m∠4 > m∠3
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7. Theorem 5.9
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8. m∠4 = m∠3 + x
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8. Definition of inequality
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9. m∠1 = m∠2 - x
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9. Subtracting equation in 8 from equation in 5
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10. m∠1 + x = m∠2
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10. Solving for m∠2
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11. m∠1 < m∠2
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11. Definition of inequality
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