5-5 Indirect Proof and Inequalities in One Triangle

Objectives

Write indirect proofs.

Apply inequalities in one triangle.

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5-5 Indirect Proof and Inequalities in One Triangle

indirect proof

Vocabulary

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5-5 Indirect Proof and Inequalities in One Triangle

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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5-5 Indirect Proof and Inequalities in One Triangle Holt Geometry

5-5 Indirect Proof and Inequalities in One Triangle Helpful Hint

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

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5-5 Indirect Proof and Inequalities in One Triangle Example 1: Writing an Indirect Proof Write an indirect proof that if a > 0, then

Step 1 Identify the conjecture to be proven.

Given: a > 0

Prove:

Step 2 Assume the opposite of the conclusion.

Assume

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5-5 Indirect Proof and Inequalities in One Triangle Example 1 Continued

Step 3 Use direct reasoning to lead to a contradiction.

Given, opposite of conclusion Zero Prop. of Mult. Prop. of Inequality

1  0 However, 1 > 0.

Simplify.

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5-5 Indirect Proof and Inequalities in One Triangle Example 1 Continued

Step 4 Conclude that the original conjecture is true.

The assumption that is false. Therefore

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5-5 Indirect Proof and Inequalities in One Triangle 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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5-5 Indirect Proof and Inequalities in One Triangle 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°.

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5-5 Indirect Proof and Inequalities in One Triangle 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.

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5-5 Indirect Proof and Inequalities in One Triangle

The positions of the longest and shortest sides of a triangle are related to the positions of the largest and smallest angles.

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5-5 Indirect Proof and Inequalities in One Triangle Example 2A: Ordering Triangle Side Lengths and Angle Measures Write the angles in order from smallest to largest.

The shortest side is , so the smallest angle is  F.

The longest side is , so the largest angle is  G.

The angles from smallest to largest are  F,  H and  G.

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5-5 Indirect Proof and Inequalities in One Triangle Example 2B: Ordering Triangle Side Lengths and Angle Measures Write the sides in order from shortest to longest.

m  R = 180° – (60° + 72°) = 48° The smallest angle is  R, so the shortest side is .

The largest angle is  Q, so the longest side is .

The sides from shortest to longest are

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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 2a Write the angles in order from smallest to largest.

The shortest side is , so the smallest angle is  B.

The longest side is , so the largest angle is  C.

The angles from smallest to largest are  B,  A, and  C.

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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 2b Write the sides in order from shortest to longest.

m  E = 180° – (90° + 22°) = 68° The smallest angle is  D, so the shortest side is .

The largest angle is  F, so the longest side is .

The sides from shortest to longest are

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5-5 Indirect Proof and Inequalities in One Triangle

A triangle is formed by three segments, but not every set of three segments can form a triangle.

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5-5 Indirect Proof and Inequalities in One Triangle

A certain relationship must exist among the lengths of three segments in order for them to form a triangle.

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5-5 Indirect Proof and Inequalities in One Triangle Example 3A: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain.

7, 10, 19

No—by the Triangle Inequality Theorem, a triangle cannot have these side lengths.

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5-5 Indirect Proof and Inequalities in One Triangle Example 3B: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain.

2.3, 3.1, 4.6

  Yes—the sum of each pair of lengths is greater than the third length.



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5-5 Indirect Proof and Inequalities in One Triangle Example 3C: Applying the Triangle Inequality Theorem Tell whether a triangle can have sides with the given lengths. Explain.

n + 6, n 2 – 1, 3n, when n = 4.

Step 1 Evaluate each expression when n = 4.

n + 6 4 + 6 10

n

2 (4) 2 – 1 15 – 1 3n 3 (4) 12

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5-5 Indirect Proof and Inequalities in One Triangle Example 3C Continued

Step 2 Compare the lengths.   Yes—the sum of each pair of lengths is greater than the third length. 

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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 3a Tell whether a triangle can have sides with the given lengths. Explain.

8, 13, 21

No—by the Triangle Inequality Theorem, a triangle cannot have these side lengths.

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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 3b Tell whether a triangle can have sides with the given lengths. Explain.

6.2, 7, 9

  Yes—the sum of each pair of lengths is greater than the third side.



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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 3c Tell whether a triangle can have sides with the given lengths. Explain.

t – 2, 4t, t 2 + 1, when t = 4

Step 1 Evaluate each expression when t = 4.

t – 2 4 – 2 2 4 4t (4) 16

t

2 (4) 2 + 1 + 1 17

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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 3c Continued

Step 2 Compare the lengths.    Yes—the sum of each pair of lengths is greater than the third length.

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5-5 Indirect Proof and Inequalities in One Triangle Example 4: Finding Side Lengths The lengths of two sides of a triangle are 8 inches and 13 inches. Find the range of possible lengths for the third side.

Let x represent the length of the third side. Then apply the Triangle Inequality Theorem.

x + 8 > 13 x > 5 x + 13 > 8 x > –5 8 + 13 > x 21 > x Combine the inequalities. So 5 < x < 21. The length of the third side is greater than 5 inches and less than 21 inches.

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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 4 The lengths of two sides of a triangle are 22 inches and 17 inches. Find the range of possible lengths for the third side.

Let x represent the length of the third side. Then apply the Triangle Inequality Theorem.

x + 22 > 17 x > –5 x + 17 > 22 x > 5 22 + 17 > x 39 > x Combine the inequalities. So 5 < x < 39. The length of the third side is greater than 5 inches and less than 39 inches.

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5-5 Indirect Proof and Inequalities in One Triangle Example 5: Travel Application The figure shows the approximate distances between cities in California. What is the range of distances from San Francisco to Oakland?

Let x be the distance from San Francisco to Oakland. x + 46 > 51 x + 51 > 46 46 + 51 > x

Δ Inequal. Thm.

x > 5 x > –5 97 > x

Subtr. Prop. of Inequal.

5 < x < 97

Combine the inequalities.

The distance from San Francisco to Oakland is greater than 5 miles and less than 97 miles.

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5-5 Indirect Proof and Inequalities in One Triangle Check It Out!

Example 5 The distance from San Marcos to Johnson City is 50 miles, and the distance from Seguin to San Marcos is 22 miles. What is the range of distances from Seguin to Johnson City?

Let x be the distance from Seguin to Johnson City. x + 22 > 50 x > 28 28 < x < 72 x + 50 > 22 x > –28 22 + 50 > x 72 > x

Combine the inequalities.

Δ Inequal. Thm.

Subtr. Prop. of Inequal.

The distance from Seguin to Johnson City is greater than 28 miles and less than 72 miles.

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