Construct the right triangle in which acute angle R has a

LEMMAS HELPING THEOREMS We will use the following theorems to help us prove the theorems found later in this sec- tion. In their role as “helping” theorems, each of the five boxed statements that follow is called a lemma. We will prove the first four lemmas, because their content is geometric. A B C Figure 3.45 B C D A 1 2 Figure 3.46 1 2 3 Figure 3.47 C A B Figure 3.48 If B is between A and C on , then and . The measure of a line segment is greater than the measure of any of its parts. See Figure 3.45. AC 7 BC AC 7 AB AC LEMMA 3.5.1 PROOF By the Segment-Addition Postulate, . According to the Ruler Postulate, meaning BC is positive; it follows that . Similarly, . These relationships follow directly from the definition of . a 7 b AC 7 BC AC 7 AB BC 7 AC = AB + BC If separates into two parts and , then and . The measure of an angle is greater than the measure of any of its parts. See Figure 3.46. m ∠ABC 7 m∠2 m ∠ABC 7 m∠1 ∠2 ∠1 ∠ABC BD LEMMA 3.5.2 PROOF By the Angle-Addition Postulate, . Using the Protractor Postulate, ; it follows that . Similarly, . m ∠ABC 7 m∠2 m ∠ABC 7 m∠1 m ∠2 7 0 m ∠ABC = m∠1 + m∠2 If is an exterior angle of a triangle and and are the nonadjacent interior angles, then and . The measure of an exterior angle of a trian- gle is greater than the measure of either nonadjacent interior angle. See Figure 3.47. m ∠3 7 m∠2 m ∠3 7 m∠1 ∠2 ∠1 ∠3 LEMMA 3.5.3 PROOF Because the measure of an exterior angle of a triangle equals the sum of measures of the two nonadjacent interior angles, . It follows that and . m ∠3 7 m∠2 m ∠3 7 m∠1 m ∠3 = m∠1 + m∠2 In , if is a right angle or an obtuse angle, then and . If a triangle contains a right or an obtuse angle, then the measure of this angle is greater than the measure of either of the remaining angles. See Figure 3.48. m ∠C 7 m∠B m ∠C 7 m∠A ∠C 䉭ABC LEMMA 3.5.4 PROOF In , . With being a right angle or an obtuse angle, ; it follows that . Then each angle and must be acute. Thus, and . The following theorem also a lemma is used in Example 1. Its proof not given depends on the definition of “is greater than,” which is found on the previous page. m ∠C 7 m∠B m ∠C 7 m∠A ∠B ∠A m ∠A + m∠B … 90° m ∠C Ú 90° ∠C m ∠A + m∠B + m∠C = 180° 䉭ABC If and , then . a + c 7 b + d c 7 d a 7 b LEMMA 3.5.5 Addition Property of Inequality The paragraph proof in Example 1 could have been written in this standard format. EXAMPLE 1 Give a paragraph proof for the following problem. See Figure 3.49. GIVEN: and PROVE: PROOF: If and , then by Lemma 3.5.5. But and by the Segment-Addition Postulate. Using substitution, it follows that . AC 7 CE CD + DE = CE AB + BC = AC AB + BC 7 CD + DE BC 7 DE AB 7 CD AC 7 CE BC 7 DE AB 7 CD A B C D E Figure 3.49 Geometry in the Real World A carpenter’s “plumb” determines the shortest distance to a horizontal line. A vertical brace provides structural support for the roof. PROOF Statements Reasons 1. and 1. Given 2. 2. Lemma 3.5.5 3. and 3. Segment-Addition Postulate 4. 4. Substitution AC 7 CE CD + DE = CE AB + BC = AC AB + BC 7 CD + DE BC 7 DE AB 7 CD The paragraph proof and the two-column proof of Example 1 are equivalent. In either format, statements must be ordered and justified. The remaining theorems are the “heart” of this section. Before studying the theorem and its proof, it is a good idea to visualize each theorem. Many statements of inequality are intuitive; that is, they are easy to believe even though they may not be easily proved. Study Theorem 3.5.6 and consider Figure 3.50, in which it appears that . m ∠C 7 m∠B Exs. 4–8 If one side of a triangle is longer than a second side, then the measure of the angle oppo- site the longer side is greater than the measure of the angle opposite the shorter side. THEOREM 3.5.6 B 4 6 A C Figure 3.50 EXAMPLE 2 Provide a paragraph proof of Theorem 3.5.6. GIVEN: , with [See Figure 3.51a.] PROVE: PROOF: Given with , we use the Ruler Postulate to locate point D on so that in Figure 3.51b. Now in the isosceles triangle BDC. By Lemma 3.5.2, ; therefore, by substitution. By Lemma 3.5.3, because is an exterior angle of . Using the two starred statements, we can conclude by the Transitive Property of Inequality that ; that is, in Figure 3.51a. m ∠B 7 m∠A m ∠ABC 7 m∠A 䉭ADB ∠5 m ∠5 7 m∠A m ∠ABC 7 m∠5 m ∠ABC 7 m∠2 m ∠2 = m∠5 CD ⬵ BC AC AC 7 BC 䉭ABC m ∠B 7 m∠A AC 7 BC 䉭ABC B A C a Figure 3.51 B A C b 3 5 4 2 1 D 쮿 쮿