1. Inequalities in Two Triangles
5-6
Inequalities in Two Triangles
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Geometry
Holt Geometry

2. Warm up: Determine if it is possible to construct a triangle with the following vertices
1. (1,1)(2,2)(3,3) 2. 3.
No- straight line
Yes
x=4 A=65 B=73 C=42
So… AB, BC, AC

3. Warm Up 4. Write the angles in order from smallest to largest.
5. The lengths of two sides of a triangle are 12 cm and 9 cm. Find the range of possible lengths for the third side.
X, Z, Y
3 cm < s < 21 cm

4. Objective
Apply inequalities in two triangles.

6. Example 1: Using the Hinge Theorem and Its Converse
Compare mBAC and mDAC.
By the Converse of the Hinge Theorem, mBAC > mDAC.

7. Example 2: Using the Hinge Theorem and Its Converse
Compare EF and FG.
Compare the sides and angles in ∆EFH angles in ∆GFH. 98
mGHF = 180° – 82° = 98°
By the Hinge Theorem, EF < GF.

8. Example 3: Using the Hinge Theorem and Its Converse
Find the range of values for k.
Step 1 Compare the side lengths in ∆MLN and ∆PLN.
LN = LN LM = LP MN > PN
By the Converse of the Hinge Theorem, mPLN < mMLN
5k – 12 < 38
Substitute the given values.
k < 10
Add 12 to both sides and divide by 5.

9. What else do we know about the mPLN
Step 2 Since PLN is in a triangle, mPLN > 0°.
5k – 12 > 0
k < 2.4
Step 3 Combine the two inequalities.
The range of values for k is 2.4 < k < 10.

10. Check It Out! Example 4
Compare mEGH and mEGF.
Compare the side lengths in ∆EGH and ∆EGF.
FG = HG EG = EG EF > EH
By the Converse of the Hinge Theorem, mEGH < mEGF.

11. Check It Out! Example 5
Compare BC and AB.
Compare the side lengths in ∆ABD and ∆CBD.
AD = DC BD = BD mADB > mBDC.
By the Hinge Theorem, BC > AB.

12. Example 6: Travel Application
John and Luke leave school at the same time. John rides his bike 3 blocks west and then 4 blocks north. Luke rides 4 blocks east and then 3 blocks at a bearing of SE. Who is farther from school? Explain.

13. Example 2 Continued
The distances of 3 blocks and 4 blocks are the same in both triangles.
The angle formed by John’s route (90º) is smaller than the angle formed by Luke’s route (100º). So Luke is farther from school than John by the Hinge Theorem.

14. PROOF #1
Write a two-column proof.
Given: C is the midpoint of BD.
m1 = m2
m3 > m4
Prove: AB > ED

15. 5.6 Indirect Proof and Inequalities in Two Triangles
Statements
Reasons
1. C is the mdpt. of BD
m3 > m4,
m1 = m2
1. Given
2. Def. of Midpoint
3. 1  2
3. Def. of  s
4. Conv. of Isoc. ∆ Thm.
5. AB > ED
5. Hinge Thm.

16. Proof #2: Proving Triangle Relationships
Write a two-column proof.
Given:
Prove: AB > CB
Proof:
Statements
Reasons
1. Given
2. Reflex. Prop. of 
3. Hinge Thm.

17. Lesson Quiz
1. Compare mABC and mDEF.
2. Compare PS and QR.
mABC > mDEF
PS < QR

18. Lesson Quiz
3. Find the range of values for z.
–3 < z < 7

19. So far you have written proofs using direct reasoning
So far you have written proofs using direct reasoning. You began with a true hypothesis and built a logical argument to show that a conclusion was true.
In an indirect proof, you begin by assuming that the conclusion is false. Then you show that this assumption leads to a contradiction. This type of proof is also called a proof by contradiction.

21. Indirect Proof #1
Write an indirect proof that a triangle cannot have two right angles.
Step 1 Identify the conjecture to be proven.
Given: A triangle’s interior angles add up to 180°.
Prove: A triangle cannot have two right angles.
Step 2 Assume the opposite of the conclusion.
An angle has two right angles.

22. Check It Out! Example 1 Continued
Step 3 Use direct reasoning to lead to a contradiction.
m1 + m2 + m3 = 180°
90° + 90° + m3 = 180°
180° + m3 = 180°
m3 = 0°
However, by the Protractor Postulate, a triangle cannot have an angle with a measure of 0°.

23. Check It Out! Example 1 Continued
Step 4 Conclude that the original conjecture is true.
The assumption that a triangle can have two right angles is false.
Therefore a triangle cannot have two right angles.