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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
Practice Test
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Exercise 23 Page 467

Notice that ∠ SQR is an exterior angle to △ QRT, so you can apply the Exterior Angle Theorem.

Statements
Reasons
1.
RQ bisects ∠ SRT
1.
Given
2.
∠ SRQ ≅ ∠ QRT
2.
Definition of angle bisector
3.
m∠ SRQ = m∠ QRT
3.
Definition of congruent angles
4.
∠ SQR is exterior to △ QRT
4.
From the diagram
5.
m∠ SQR > m∠ QRT
5.
Exterior Angle Theorem
6.
m∠ SQR > m∠ SRQ
6.
Substitution

Since RQ bisects ∠ SRT we have that ∠ SRQ ≅ ∠ QRT, which means that m ∠ SRQ = m ∠ QRT.

Next, we notice that ∠ SQR is an exterior angle to △ QRT.

Therefore, by applying the Exterior Angle Theorem, we have that m ∠ SQR > m ∠ QRT but, since m ∠ SRQ = m ∠ QRT, we obtain that m ∠ SQR > m ∠ SRQ.

Two-Column Proof

In the following table we summarize the proof we did before.

Statements
Reasons
1.
RQ bisects ∠ SRT
1.
Given
2.
∠ SRQ ≅ ∠ QRT
2.
Definition of angle bisector
3.
m∠ SRQ = m∠ QRT
3.
Definition of congruent angles
4.
∠ SQR is exterior to △ QRT
4.
From the diagram
5.
m∠ SQR > m∠ QRT
5.
Exterior Angle Theorem
6.
m∠ SQR > m∠ SRQ
6.
Substitution
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Exercises 2 (Page 467)
Exercises 3 (Page 467)
Exercises 4 (Page 467)
Exercises 5 (Page 467)
Exercises 6 (Page 467)
Exercises 7 (Page 467)
Exercises 8 (Page 467)
Exercises 9 (Page 467)
Exercises 10 (Page 467)
Exercises 11 (Page 467)
Exercises 12 (Page 467)
Exercises 13 (Page 467)
Exercises 14 (Page 467)
Exercises 15 (Page 467)
Exercises 16 (Page 467)
Exercises 17 (Page 467)
Exercises 18 (Page 467)
Exercises 19 (Page 467)
Exercises 20 (Page 467)
Exercises 21 (Page 467)
Exercises 22 (Page 467)
Exercises 23 (Page 467)
Exercises 24 (Page 467)
Exercises 25 (Page 467)