THEOREM VIII. Hypothesis." When two sides of a triangle are equal,” THEOREM IX. Hypothesis." When two angles of a triangle are equal," Conclusion "The sides opposite to them are equal.” The hypothesis of Theorem VIII. is the conclusion of Theorem IX., and the conclusion of Theorem VIII. is the hypothesis of Theorem IX. Converse Propositions.-When there are two propositions, such that the hypothesis of the one is the conclusion of the other and the conclusion of the one the hypothesis of the other, each proposition is said to be the converse of the other. We will take another example of two propositions which have this relation to one another : Construct a triangle, A B C, having its sides unequal, so that AC, for instance, is longer than A B (Fig. 54), then it will be found that the angle B, which is opposite to the longer side, A C, will be greater than the angle C, which is opposite to the shorter side, A B. B A Again, draw a line, B C, at C, make a small angle, A C B, and at B a larger angle, A B C, so that the sides of the angles form a triangle, ABC; it will then be found that the side, A C, which is opposite the greater angle, B, is greater than the side, A B, which is opposite to the smaller angle, C. The two statements, of which these are illustrations, are as follow: FIG. 54. THEOREM X. Hypothesis." When in a triangle two sides are unequal," Conclusion "The angle opposite the greater side is greater than the angle opposite the less." THEOREM XI. (The converse of Theorem X.) Hypothesis." When two angles of a triangle are unequal," Conclusion "The side opposite the greater angle is greater than the side opposite to the less." One of two converse propositions may be true while the other is false; hence we must be careful not to assume the truth of a proposition simply because the converse has been proved. For example, it is true that any two right angles are equal to one another; but the converse of this statement is not true, namely, that any two angles which are equal to one another are right angles. It frequently happens that the enunciations of converse propositions are so given that their relation is not at once evident. This is the case with the converse Theorems I. and II. The first may be stated thus: When the exterior sides of two adjacent angles are in the same straight line, the angles are supplementary. When so stated, its relation to Theorem II. is apparent. Applications of Symmetrical Triangles.—Architects, in designing buildings, usually endeavour to make the parts symmetrical, so that oblique lines on one side correspond to oblique lines on the other side. Hence the symmetrical or isosceles triangle is frequently formed in the construction of the roof of a building. Thus a trussed beam is an isosceles triangle. Fig. 55a represents half such a beam. TT is termed the tie-beam; PP, the king-post; A A, the principal rafter or "principal;" ff, the strut. The pediment in architecture and the gable in ordinary dwellinghouses are isosceles triangles (Fig. 556.) Other Properties of the Isosceles Triangle.-We will now use the trussed beam to illustrate some important properties of the isosceles triangle, which form the subject of the following Theorems for Exercise. Let us suppose that the two equal FIG. 556. principals, the tie-beam and the king-post, are made and have to be fitted together. They may be put together in several different ways. First way. If the principals are fitted to the tie-beam, and the king-post so placed that it makes equal angles with the principals, then the king-post must fall into its place at the middle of the tie-beam and be perpendicular to the tie-beam.-Exercise 9, p. 43. Second way.If the head of the king-post be attached to the two principals and its foot driven into its position in the middle of the tiebeam, then the king-post must necessarily be perpendicular to the tiebeam, and must make equal angles with the principals.-Exercise 10, P. 43. Third way. If the head of the king-post be attached to the two principals, and the king-post be placed so that it is perpendicular to the tie-beam, its foot must necessarily fall into its position at the middle of the tie-beam, and it must make equal angles with the principals.— Exercise 11, p. 45. Easy Exercises. (a). With the point A as centre, and with a radius of one inch, describe a circle; and then, with B as centre, describe another circle equal to the first. (b). Call the points, in which these circles cut one another, C and D. Join the points C and D, and measure the line C D. Let the point in which C D and A B intersect be called E. Measure AE and B E. What angles are formed at the point E? (c.) Take a line, A B, and find its middle point, C. With AB as radius and B as centre, describe a circle; and, with A C as radius and C as centre, describe another circle. Take any point you please on the larger circumference, and call it M. If a small fly walk from A to M, how much of its path will be described when it reaches the circumference of the smaller circle? (d). Make an equilateral triangle, each side of which shall be one inch long. (e). Divide a sheet of paper into equilateral triangles. THEOREMS ON CHAPTER IV. SECTION II. 41 DEFINITIONS. 32. An isosceles triangle is a triangle which has two sides equal. 33. An equilateral triangle is a triangle which has all its sides equal. 34. A right-angled triangle is a triangle one of whose angles is a right angle. The side opposite the right angle in a right-angled triangle is termed the hypothenuse. THEOREM VIII. When a triangle has two sides equal, the angles opposite to these sides are equal, or the angles at the base of an isosceles triangle are equal to one another. == Let A B C be a triangle in which AB A C, then shall ▲ B=▲ C. Let A D be the straight line which bisects A and meets B C in the point D. Then the triangles ABD, ACD have A B and AD equal respectively to AC and AD, and = A B D C < BAD 4 CAD; therefore the triangles are equal in all respects, and ▲ B= 4 C. THEOREM IX. When a triangle has two angles equal the sides opposite these angles are equal. Let A B C be a triangle in which BC; then shall A B AC. B Suppose B D to be a straight line, making with B C an angle CBD equal to angle C B A, but on the opposite side of B C. Also suppose CD to make with B C an angle B CD, equal to B C A, but on the opposite side of B C. Then, because ▲ BCA = ▲ CBA, therefore BCA = 4 CBD, and CBA 4 BCD. Since the triangles A B C, D C B, have ABC, A C B, and side B C equal respectively to ▲ DC B, DB C, and side BC; therefore the triangles are equal, and the sides opposite the equal angles are equal. Consequently AC = DB. But the triangles also have ▲ ABC, ▲ ACB, and side BC respectively equal to 4 DBC, 4 D C B, and side BC, and therefore the triangles are equal, and the sides opposite the equal angles are equalConsequently AB = DB; therefore AB = AC. A THEOREM X. When two sides of a triangle are unequal the angle opposite the greater side is greater than the angle opposite the less. Let A B C be a triangle, having the side A C greater than the side A B, then shall ABC be greater than 4 A CB. From AC cut off a part A D equal to A B, and draw the line B D. B D Then because A B=AD, therefore LABD= LAD B. = Hence A B C is greater than A D B. And because ADB is the exterior angle of ▲ BD C. Therefore A D B is greater than ▲ A CB, Consequently ▲ ABC must be greater than 4 A CB. |