3rd Syllogism. If two triangles have two sides equal to two sides, and the base of one equal to the base of the other, those sides shall contain equal angles. (Euc. I. 8.) The triangles AD Fand A E F have two sides A D, A F equal to two sides A E, AF(); and the base D Fequal to the base E F (d). ... The angle DAF equals the angle E AF, Result.—Wherefore the angle B A C has been divided into the two equal angles B A F, CA F. Note.—According to what was said on the definition of an angle (page 27), angle B A F is the same as D AF; and angle CA F is the same as E A F. Proof (with contracted syllogisms, in the words of Euclid). Because A D is equal to A E (by construction), and A F is common to the triangles ADF, Á EF. ... The two sides DA, A Fare equal to the two sides E A, A F, each to each. And the base DF is equal to the base EF. (Definition.) .. The angle DAF equals the angle EAF. (Euc. I. 8.) Result.-Wherefore the given rectilineal angle BAC is bisected by the line A F. Q. E, F, EXERCISE. Bisect the given rectilineal angle 1, 2, 3, using figures instead of letters to describe the lines, &c. А PROBLEM (Euclid 1. 10.) Repeat.— The definition of an equilateral triangle. (Euc. I. 4.-If two triangles have two sides of the one, equal to two sides of the other, each to each ; and have also the angle contained by the two sides of the one equal to the angle contained by the two sides of the other ; then the bases shall be equal. (Assumed here as an axiom proved on page 70.) General Enunciation. To bisect a given finite straight line-that is, to divide it into two equal parts. Particular Enunciation. Given.—The finite straight B line A B. Required.—To divide A B into two equal parts. Construction. (a) On A B describe an equilateral triangle ABC (Euc. I. I shows how.) (6) Bisect the angle' A CB by the straight line CD, meeting AB at D. (Euc. I. 9 shows how to do this.) Then the line A B shall be bisected at the point D. If A B be bisected at D, then we have to prove that A D is equal to D B. Proof (with syllogisms in full.) CAD and CBD are two triangles (distinguish between angles and triangles). А. AC, CD are two sides of triangle CAD, and ACD is the angle contained by them. BC, CD are two sides of triangle CBD, and BCD is the angle contained by them. ist Syllogism. C A is equal to CB (by Construction a); add C D to each of these. (c) Then A C, C D are equal to BC, CD, each to each. 2nd Syllogism. If two triangles have two sides equal to two sides, and the included angle of one equal to the included angle of the other, the bases of those triangles shall be equal. The triangles CAD and CBD have two sides AC, CD equal to two sides BC, CD), and they have the included angle A CD, equal to the included angle BCD (Con struction b.) ... The base A D is equal to the base D B. Result.-Wherefore the given straight line AB is bisected at the point D. Proof (with contracted syllogisms). Because AC is equal to CB (definition of equilateral triangle); and C D is common to the two triangles A CD, BCD. ... The two sides A C, C D are equal to the two sides BC, CD. And the angle A CD is equal to the angle B CD (by construction). .'. The base A D is equal to the base D B. (Euc. I. 4.) Result.—Wherefore the given line A B is divided into two equal parts at D. Q. E. F. D EXERCISES.-I. In the given straight line A B find a point that shall be equally distant from C and D. (Use an equilateral triangle, and bisect the angle at the vertex.-) II. If straight lines are drawn C from the points A and B to any point in the line CD, those lines are found to be equal, and the angle thus made is found to be bisected by the line CD. Prove that A and B are at equal distances from the line CD. (Draw lines to any point in CD and join A B.) A B DEFINITION OF RIGHT ANGLES. Euc. Def. 10.) — When a straight line, standing on another straight line, makes the adjacent angles equal to Fig.1 one another, each of the angles is called a right angle. And the straight line which stands on the other is A said to be perpendicular to it. Thus the straight line CD is said to stand on the straight line A B, and to be perpendicular to it. It is evident that if one straight line stand on · The vertex of a triangle is the angle opposite to the base. D B another it forms two angles. In the figures given here they are the angles CDA, CDB. Now those angles must be either equal or unequal. In fig. 1 they are equal. Fig.2 In fig. 2 they are not right angles. In fig. 1 the straight line C D may be drawn either from D, a point in A B, or from C, a point without (outside) A B. If drawn from D, the line CD is said to be at right angles to A B. If drawn from C, the line CD is said to be perpendicular to A B. A D PROBLEM (Euclid I. 11). Repeat.—The definition of a right angle, and of an equilateral triangle, and Axiom II. (Euc. I. 8.)—If two triangles have two sides of the one equal to two sides of the other, each to each; and have also their bases equal; then the angle contained by the two sides of one triangle shall be equal to the angle contained by the two sides of the other. (Assumed as an axiom, proved on page 76.) General Enunciation. To draw a straight line at right angles to a given straight line, from a given point in the same. |