Download - Holt Geometry 5-6 Inequalities in Two Triangles 5-6 Inequalities in Two Triangles Holt Geometry Warm Up Warm Up Lesson Presentation Lesson Presentation.

Holt Geometry

5-6 Inequalities in Two Triangles5-6 Inequalities in Two Triangles

Holt Geometry

Warm Up

Lesson Presentation

Lesson Quiz

Holt Geometry

5-6 Inequalities in Two Triangles

Warm Up

1. Write the angles in order from smallest to largest.

2. 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

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Apply inequalities in two triangles.

Write an indirect proof.

Objective

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Example 1A: Using the Hinge Theorem and Its Converse

Compare mBAC and mDAC.

Compare the side lengths in ∆ABC and ∆ADC.

By the Converse of the Hinge Theorem, mBAC > mDAC.

AB = AD AC = AC BC > DC

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Example 1B: Using the Hinge Theorem and Its Converse

Compare EF and FG.

By the Hinge Theorem, EF < GF.

Compare the sides and angles in ∆EFH angles in ∆GFH.

EH = GH FH = FH mEHF > mGHF

mGHF = 180° – 82° = 98°

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Example 1C: Using the Hinge Theorem and Its Converse

Find the range of values for k.

Step 1 Compare the side lengths in ∆MLN and ∆PLN.

By the Converse of the Hinge Theorem, mMLN > mPLN.

LN = LN LM = LP MN > PN

5k – 12 < 38

k < 10

Substitute the given values.

Add 12 to both sides and divide by 5.

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Example 1C Continued

Step 2 Since PLN is in a triangle, mPLN > 0°.

Step 3 Combine the two inequalities.

The range of values for k is 2.4 < k < 10.

5k – 12 > 0

k < 2.4

Substitute the given values.

Add 12 to both sides and divide by 5.

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Check It Out! Example 1a

Compare mEGH and mEGF.

Compare the side lengths in ∆EGH and ∆EGF.

FG = HG EG = EG EF > EH

By the Converse of the Hinge Theorem, mEGH < mEGF.

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Check It Out! Example 1b

Compare BC and AB.

Compare the side lengths in ∆ABD and ∆CBD.

By the Hinge Theorem, BC > AB.

AD = DC BD = BD mADB > mBDC.

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Example 2: 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 N 10º E. Who is farther from school? Explain.

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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.

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Example 3: Proving Triangle Relationships

Write a two-column proof.

Given:

Prove: AB > CB

Proof:

Statements Reasons

1. Given

2. Reflex. Prop. of

3. Hinge Thm.

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Check It Out! Example 3a


Given: C is the midpoint of BD.

Prove: AB > ED

m1 = m2

m3 > m4

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1. Given

2. Def. of Midpoint

3. Def. of s

4. Conv. of Isoc. ∆ Thm.

5. Hinge Thm.

1. C is the mdpt. of BDm3 > m4, m1 = m2

3. 1 2

5. AB > ED

Statements Reasons

Proof:

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Given:

Prove: mTSU > mRSU

Statements Reasons

1. Given

3. Reflex. Prop. of

4. Conv. of Hinge Thm.

2. Conv. of Isoc. Δ Thm.

1. SRT STRTU > RU

SRT STRTU > RU

Check It Out! Example 3b

4. mTSU > mRSU

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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.

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When writing an indirect proof, look for a contradiction of one of the following: the given information, a definition, a postulate, or atheorem.

Helpful Hint

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Example 1: Writing an Indirect Proof

Step 1 Identify the conjecture to be proven.

Given: a > 0

Step 2 Assume the opposite of the conclusion.

Write an indirect proof that if a > 0, then

Prove:

Assume

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Example 1 Continued

Step 3 Use direct reasoning to lead to a contradiction.

However, 1 > 0.

1 0

Given, opposite of conclusion

Zero Prop. of Mult. Prop. of Inequality

Simplify.

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Step 4 Conclude that the original conjecture is true.

Example 1 Continued

The assumption that is false.

Therefore

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Check It Out! Example 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.

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Check It Out! Example 1 Continued

Step 3 Use direct reasoning to lead to a contradiction.

However, by the Protractor Postulate, a triangle cannot have an angle with a measure of 0°.

m1 + m2 + m3 = 180°

90° + 90° + m3 = 180°

180° + m3 = 180°

m3 = 0°

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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.

Check It Out! Example 1 Continued

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Lesson Quiz: Part I

1. Compare mABC and mDEF.

2. Compare PS and QR.

mABC > mDEF

PS < QR

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Lesson Quiz: Part II

3. Find the range of values for z.

–3 < z < 7

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Statements Reasons

1. Given

2. Reflex. Prop. of

3. Conv. of Hinge Thm.3. mXYW < mZWY

Given:

Prove: mXYW < mZWY

4. Write a two-column proof.

Lesson Quiz: Part III

Proof: