STRAIGHT LINES AND RECTILINEAL PLANE FIGURES.* 23. PROPOSITION I. To describe an equilateral triangle upon a given straight line↑. Let AB be the given straight line, which is to be one side of the triangle; with centre A and radius AB (POST. III. 20) trace a portion of the circumference of a circle on that side of AB on which the triangle is required; with the same radius and with centre B trace another portion of the circumference of a circle on the same side of AB, and intersecting the former in the point C; join the points A and C by the straight line AC (POST. 1.), and B and C by the straight line BC; then ABC shall be the equilateral triangle required. = B For since B and C are points in the circumference of the same circle whose centre is A, AB AC, (Def. 16); again, since A and C are points in the circumference of the same circle whose centre is B, AB or BA = BC ; .. AC=AB= BC, or the three sides of the triangle ABC are equal to each other, that is, ABC is an equilateral triangle and it is described upon the straight line AB. 24. PROP. II. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles formed by those sides equal to one another, they shall also have their bases, or third sides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite. Let ABC, DEF be two triangles, in which the side A rectilineal plane figure means a plane surface (7) bounded by straight lines. According to the number of such lines, forming its boundary, each figure receives its distinctive name. The Author does not deem it advisable to deviate much from Euclid's mode of expression, but rather to explain it, when it appears necessary, in a note. Thus, in this instance, to describe' a triangle means to construct or trace it; and upon a given straight line' means so as to have that straight line for its BASE. Also a given straight line' means here a line fixed both in position and magnitude. 6 manner that the point A upon the point D, and the line AB upon DE; then the point B will fall B upon E, because AB = DE. Again, since AB falls upon DE, AC will also fall upon DF, because <BAC=¿EDF, (8.) Since, then, the point A is upon the point D, and the line AC upon DE, the point C shall fall upon F, because ACDF. Hence, since B is upon E, and C upon F, the line BC must coincide with EF, because BC and EF are straight lines between the same, or coincident, points. Therefore the triangles coincide, and are equal, in all respects, as stated above. COR. Hence, also, if two triangles have the three sides of the one equal to the three sides of the other, each to each, in the same order, the two triangles will be equal, and their angles likewise will be equal, each to each, viz. those to which the equal sides are opposite. For it is evident from what has been shewn above, that such triangles, applied to each other as in the former case, will coincide in every part, and therefore be equal in all respects*. 25. PROP. III. To bisect a given angle, that is, to divide it into two equal angles. Let BAC be the given angle; it is required to bisect it. In AB take any point D, and with centre A and radius AD describe an arc of a circle cutting AC in the point E; join the points D, and E, by the straight line DE, and upon DE describe the equilate- A E B * EUCLID does not seem to have considered this sufficiently evident, and therefore proves it by the process, usually called reductio ad absurdum, before explained. + Given, that is, by being traced on a given plane surface. ral triangle DEF (23); then join AF, and the angle BAC is bisected by the line AF. For, since D and E are points in the circumference of the same circle whose centre is A, AD = AE; and since DEF is an equilateral triangle, DF = EF. Therefore in the two triangles ADF, AEF, the three sides AD, DF, AF are equal to the three sides AE, EF, AF, each to each, in order; so that (24, COR.) the two triangles are equal in all respects, and the angle DAF between AD, AF, is equal to the angle EAF between AE, AF. Therefore the angle BAC is divided into two equal angles by the straight line AF. 26. PROP. IV. The angles at the base* of an isosceles triangle are equal to one another. Let ABC be an isosceles triangle, in which the side AB= the side AC, and BC is the third side, or base; the angle ABC shall be equal to the angle ACB. = Bisect the angle BAC by the straight line AD (25), D being the point where AD meets the base BC. Then, since AB AC, and ▲ BAD = L CAD, we have two triangles ABD, ACD, in which the two sides BA, AD are equal to the two sides CA, AD, each to each, and the angle formed by the two sides of the one equal to the angle formed by the two sides of the other, .. the triangles are equal in all respects (24), and the angles are equal, each to each, to which the equal sides are opposite; and or, .. LABDL ACD, which is the same thing, ▲ ABC = ‹ ACB. D COR. Hence every equilateral triangle is also equiangular; and, conversely, every equiangular triangle is also equilateral. 27. PROP. V. To bisect a given finite straight line, that is, to divide it into two equal straight lines. *The base in an isosceles triangle is restricted to one side, viz. the unequal side, on which the two equal sides may be supposed to stand, except when the triangle is also equilateral, in which case any side may be taken as the base. A given finite straight line' means a straight line fixed both in position and magnitude. Let AB be the given straight line. Upon AB describe the equilateral triangle ABC (23); bisect the angle ACB by the straight line CD meeting AB in D (25); then AB is bisected in the point D. For AC, CD are equal to BC, CD, each to each, and LACDL BCD,.. the triangles ACD, BCD are equal in all respects (24); and .. AD BD, B being the sides opposite to the equal angles ACD, BCD; that is, AB is divided into two equal parts in the point D. 28. PROP. VI. To draw a straight line at right angles to a given straight line* from a given point in it. Let AB be the given straight line, and C a given point in it. It is required to draw a straight line from C at right angles to AB. In AC take any point D, and with centre C, and radius CD, describe an arc of a circle A D E B cutting the line AB in D and E; upon DE describe the equilateral triangle DEF (23); and join FC; CF shall be at right angles to AB. For, since CD = CE (16), and DF = EF (23), by construction, in the two triangles DCF, ECF, DC, CF are equal to EC, CF, each to each, and the third side DF is equal to EF, .. the_triangles are equal in all respects, (24, Cor.) and 4DCF = ▲ ECF, to which the equal sides DF, EF are respectively opposite, and.. each of them is a right angle (10), that is, CF is at right angles to AB. 29. PROP. VII. To draw a straight line † perpendi This straight line is required to be given in position only. Whether a certain straight line is drawn at right angles, or perpendicular, to another straight line, depends upon the simple fact, whether it be drawn, from a known point in the latter line itself, away from the line, or from a known point without it towards the line. cular to a given straight line of unlimited length from a given point without it. Let AB be the given straight line, and C a given point without it, from which it is required to draw a perpendicular to AB. Take a point D on the other side of AB, and with centre C and radius CD, describe a circle cutting AB, or AB produced, in E and G D F; join CE, CF, and bisect ECF by the line CG, meeting AB in G. Then CG shall be perpendicular to AB. For CE CF (16), and 4 ECG = <FCG, by construction, .. in the triangles ECG, FCG, EC, CG are equal to FC, CG, each to each, and ECG = < FCG, ..the triangles are equal in all respects (24), and the angles equal to which equal sides are opposite, viz. LEGC=L FGC, and.. each of them is a right angle, or CG is perpendicular to AB (10). 30. PROP. VIII. The two angles which one straight line makes with another upon the one side of it are always equal to two right angles. Let the straight line CD meet the straight line AB in the point C, and make with AB on the one side of it the angles ACD, BCD; these are always equal to two right angles. 2 For, if ACD = 4 BCD, then each of them is a right angle (10), and .. the two together make two right angles. A D Or, if the angles ACD, BCD are unequal, from the point Cin AB draw CE at right angles to AB (28); then, supposing ACD to be the greater of the two angles ACD, BCD, it is evident that ACD is as much greater than a right angle as BCD is less, and .. that the two together are equal to two right angles. L |