HISTORICAL NOTE. The Egyptians showed no knowledge of a logical demonstration, nor did the Arabians, who studied geometry quite extensively. The Greeks developed the process of demonstration to a high state of perfection. They were fully aware, moreover, that certain propositions must be admitted without proof (see § 60). Thus Aristotle (384-322 B.C.) says: "Every demonstration must start from undemonstrable principles. Otherwise the steps of a demonstration would be endless." Euclid divided unproved propositions into two classes: axioms, or "common notions," which are true of all things, such as, "If things are equal to the same thing they are equal to each other"; and postulates, which apply only to geometry, such as, "Two points determine a line." The best usage in modern mathematics is to adopt the one word axiom for both of these, as in § 60. Much practice is needed in writing demonstrations in full detail. This should be done in the shortest possible sentences, usually giving a separate line to each statement, followed by the definition, axiom, or theorem on which it depends. For this purpose the following symbols and abbrevia INEQUALITIES OF PARTS OF TRIANGLES. 81. Definition. If one side of a triangle is produced, the angle thus formed is called an exterior angle of the triangle. Thus, 21 is an exterior angle of the triangle ABC. A 1 B 82. Axiom VII. If a, b, c are line-segments (or angles) such that a>b and bc, or such that ab and b>c, then a> c. The proof of the following theorem is shown in full detail as it should be written by the pupil or given orally, except that the numbers of paragraphs should not be required. 83. THEOREM. An exterior angle of a triangle is greater than either of the opposite interior angles. Given the ▲ ABC with the exterior angle DBC formed by pro ducing the side AB. To prove that DBC > < C and also DBC > ▲ A. Proof: Let E be the middle point of BC. Find E by the construction for bisecting a line-segment (§ 51). Draw AE and prolong it, making EF = AE, and draw BF. In the two A ACE and FBE, we have by construction CE = EB and AE = EF. Also LCEA L BEF. = (Vertical angles are equal, § 74.) ... A ACE AFBE. (Two triangles which have two sides and the included angle of the one equal respectively to two sides and the included angle of the other are congruent, § 32.) ... LC = LFBE. (Being angles opposite equal sides in congruent triangles, § 29.) (If an angle is the sum of two angles it is greater than either of them, § 39.) ... < DBC > < c. (Since DBC> <FBE and ▲ FBE = ≤ C, Ax. VII, § 82.) In order to prove ≤ DBC >▲ A, prolong CB to some point G.. Then LABG = DBC. (Vertical angles are equal, § 74.) Now bisect AB, and in the same manner as before we may prove LABGA. .. ≤ DBC > LA. (Since DBC = ▲ ABG and ▲ ABG>Z A, Ax. VII, § 82.) For the second part of the proof let I be the middle point of AB. Draw CH and prolong it to K, making CH = HK. Let the student draw the figure for the second part of the proof and give it in full. Hereafter more and more of the details of the proofs will be left for the student to fill in. When reference is made to a paragraph in the text or when the reason for a step is called for, the complete statement of the definition, axiom, or theorem should be given by the student. 84. THEOREM. If two sides of a triangle are unequal, the angles opposite these sides are unequal, the greater angle being opposite the greater side. 85. THEOREM. If two angles of a triangle are unequal, the sides opposite them are unequal, the greater side being opposite the greater angle. Given AABC in which <B>LA. Proof: One of the following three statements must be true: (1) b=a, (2) b<a, (3) b>a. But it cannot be true that b =α, for in that case LB=LA, A contrary to the hypothesis that ≤ B > < A. b a And it cannot be true that b < a, for in that case B (§ 37) LB < LA, contrary to the hypothesis. Hence it follows that b> a. (§ 84) 86. The above argument is called proof by exclusion. Its success depends upon being able to enumerate all the possible cases, and then to exclude all but one of them by showing that each in turn leads to some contradiction. 1. The hypotenuse of a right triangle is greater than either leg. 2. Show that not more than two equal line-segments can be drawn from a point to a straight line. SUGGESTION. Suppose a third drawn. Then apply §§ 37, 83, 84. 3. Show by joining the vertex A of the triangle ABC to any point of the side BC that 4B + ≤ C < 2 rt. 4. Use § 83. 4. If two angles of a triangle are equal, the sides opposite them are equal. Use §§ 84, 86. 5. Either leg of an isosceles triangle is greater than half of the base. 6. Show that an equiangular triangle is equilateral, and conversely. THEOREMS ON PARALLEL LINES. 88. A straight line which cuts two straight lines is called a transversal. The various angles formed are named as follows: 24 and 25 are alternate-interior angles; also 3 and ≤6. 4/3 6/5 8/7 22 and 27 are alternate-exterior angles; also <1 and 48. 21 and 25 are corresponding angles; also ≤3 and ≤7, 42 and 26, 4 and 8. ≤3 and ≤5 are interior angles on the same side of the transversal; also 24 and 26. 89. Definition. Two complete lines which lie in the same plane and which do not meet are said to be parallel. Two line-segments are parallel if they lie on parallel lines. |