To construct a regular octagon, what measure would be To construct a regular dodecagon 12 sides, what

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