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If four lines go out from a point so as to make each angle equal to the one not adjacent to it, the four lines will form only two intersecting lines. HYPOTHESIS. Let OA, OB, OC, OD, be four lines, with the common point O; and let * AOB X COD, and X BOC = * DOA CONCLUSIONS. AO and OC are in one line. BO and OD are in one line. PROOF. | AOB + % BOC + 4 COD + 4 DOA = a perigon, (71. The sum of all the angles about a point in a plane is a perigon.) By hypothesis, % AOB = X COD, and 4 BOC = 4 DOA, .. twice X AOB + twice X BOC = a perigon; * AOB + X BOC : = a st. %. (107. Half a perigon is a straight angle.) .:. AO and OC form one line. (65. If two supplemental angles be added, their exterior arms will form one line.) In the same way we may prove that X AOB and X DOA are supplemental adjacent angles, and so that BO and OD form one line. 110. COROLLARY. The four bisectors of the four angles formed by two lines intersecting, form a pair of lines perpendicular to each other. II. Angles about Two Points. III. A line cutting across other lines is called a Transversal. 112. If in a plane two lines are cut in two distinct points by a transversal, at each of the points four angles are determined. Of these eight angles, four are between the two given lines (namely, 4, 3, a, 6), and are called Interior Angles; the other four lie outside the two lines, and are called Exterior Angles. Angles, one at each point, which lie on the same side of the transversal, the one exterior, the other interior, like I and a, are called Corresponding Angles. Two angles on opposite sides of the transversal, and both interior or both exterior, like 3 and a, are called Alternate Angles. THEOREM V. 113. If two corresponding or two alternate angles are equal, or if two interior or two exterior angles on the same side of the transversal are supplemental, then every angle is equal to its corresponding and to its alternate angle, and is supplemental to the angle on the same side of the transversal which is interior or exterior according as the first is interior or exterior, FIRST CASE. HYPOTHESIS. Xa= *1. CONCLUSIONS. Xa= *1 = 3 = X 6. X 2 b = 72 = 44 = X d. Ya+*4 = 41+ d = 4 2 + %= %6+43 = st. %. 2/1 3 / 4 z la cld PROOF. Xa=X.c, 2.b = X d, 41 = 43, 4.2 = * 4. (108. If two lines intersect, the vertical angles are equal.) Moreover, since, by hypothesis, xa = % 1, their supplements are equal, or Xb= x 2. (98. All straight angles are equal.) (89. If equals be taken from equals, the remainders are equal.) Xa = XI= *C= X3, X5 X 2 = *4, and st. * xatxb a to * 4 = *It * d = * It * d = 4 2 + 0 Xb+ X 3. (88. If equals be added to equals, the sums are equal.) SECOND CASE. If, instead of two corresponding, we have given two alternate angles equal, we substitute for one its vertical, which gives the First Case again. THIRD CASE. HYPOTHESIS. Xa + 4.4 = st. 4. But * 4+ X 1 = st. 4, i. Xa = 1, which gives again the First Case. FOURTH CASE. — HYPOTHESIS. 41+4 d = st. %. But at * d = st. 4, 1. Xa = x 1, which gives again the First Case. III. Triangles. 114. An Equilateral Triangle is one in which the three sides are equal. Δ Δ 115. An Isosceles Triangle is one which has two sides equal. 116. A Scalene Triangle has no two sides equal. 117. When one side of a triangle has to be distinguished from the other two, it may be called the Base; then that one of the vertices opposite the base is called the Vertex. 118. When we speak of the angles of a triangle, we mean the three interior angles. 119. A Right-angled Triangle has one of its angles a right angle. The side opposite the right angle is called the Hypothe nuse. 120. An Obtuse-angled Triangle has one of its angles obtuse. 121. An Acute-angled Triangle has all three angles acute. 122. An Equiangular Triangle is one which has all three angles equal. 123. When two triangles have three angles of the one equal respectively to the three angles of the other, a pair of equal angles are called Homologous Angles. The pair of sides opposite homologous angles are called Homologous Sides. THEOREM VI. 124. Two triangles are congruent if two sides and the included angle in the one are equal respectively to two sides and the included angle in the other. M B BC= I 케 a HYPOTHESIS. ABC and LMN two triangles, with AB = LM, MN, * B = * M. CONCLUSION. The two triangles are congruent; or, using a for (triangle," and = for “congruent," A ABCALMN. PROOF. Apply the triangle LMN to A ABC in such a manner that the vertex M shall rest on B, and the side ML on BA, and the point N on the same side of BA as C. Then, because the side ML equals BA, the point I will rest upon A; because X M = * B, the side MN will rest upon the line BC; because the side MN equals BC, the point N will rest upon the point C. Now, since the point L rests upon A, and the point N rests upon C, therefore the side LN coincides with the side AC. (95. If two lines have two points in common, the two sects between those points coincide.) Therefore every part of one triangle will coincide with the corresponding part of the other, and the two are congruent. (94. Magnitudes which will coincide are congruent.) 125. In any pair of congruent triangles, the homologous sides are equal. « AnteriorContinuar »
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