DC. Then, by adding BC to both sides we get AB + BC > BC+ DC, and by the Segment Addition Postulate we obtain AC > BD. AF≅ '>
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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Inequalities in Two Triangles
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Exercise 26 Page 459

Add BC to both sides of the given inequality and use the Segment Addition Postulate. Next, apply the Converse of the Hinge Theorem.

Statements
Reasons
1.
AB> CD
1.
Given
2.
AB +BC > CD + BC
2.
Adding BC to both sides
3.
AC = AB +BC and BD = CD + BC
3.
Segment Addition Postulate
4.
AC > BD
4.
Substitution
5.
AF≅ DJ and FC≅ JB
5.
Given
6.
m∠ AFC > m∠ DJB
6.
Converse of the Hinge Theorem
Practice makes perfect

Let's highlight the congruent segments AF≅ DJ and FC≅ JB in the diagram.

We are told that AB > DC. Then, by adding BC to both sides we get AB + BC > BC+ DC, and by the Segment Addition Postulate we obtain AC > BD. AF≅ DJ FC≅ JB AC > BD Finally, by the Converse of the Hinge Theorem we get that m∠ AFC > m∠ DJB, which is what we wanted to prove.

Two-Column Proof

In the following table we will summarize the proof we did above.

Statements
Reasons
1.
AB> CD
1.
Given
2.
AB +BC > CD + BC
2.
Adding BC to both sides
3.
AC = AB +BC and BD = CD + BC
3.
Segment Addition Postulate
4.
AC > BD
4.
Substitution
5.
AF≅ DJ and FC≅ JB
5.
Given
6.
m∠ AFC > m∠ DJB
6.
Converse of the Hinge Theorem
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arrow_right Exercises p.457-462
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Exercises 2 (Page 457)
Exercises 3 (Page 457)
Exercises 4 (Page 457)
Exercises 5 (Page 458)
Exercises 6 (Page 458)
Exercises 7 (Page 458)
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Exercises 17 (Page 459)
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Exercises 28 (Page 460) engineering
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Exercises 39 (Page 461) engineering
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