into two equal parts. Then, in the triangles BAF, CAF, the side BA is equal (hyp.) to CA, and AF is common; therefore the two sides BA, AF, are equal to the two CA, AF, each to each, and the contained angles BAF, CAF are equal: wherefore (I. 4, part. 3) the remaining angles are equal, each to each, to which the equal sides are opposite; therefore, the angle ABF is equal to the angle ACF, and (I. 4, schol.) the exterior angles DBF, ECF, are D also equal. B P C E Cor. Hence every equilateral triangle is also equiangular. This is shown by taking, first, one side as base, and then another. PROP. VI. THEOR.*If two angles of a triangle be equal to one another, the sides which subtend, or are opposite to, those angles, are also equal to one another. Let ABC be a triangle having the angle ABC equal to the angle ACB; the side AB is also equal to the side AC. D Λ For, if AB be not equal to AC,† one of them is greater than the other let AB be the greater, and from it cut off (I. 3) BD equal to AC, the less, and join DC. Then, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each; and (hyp.) the angle DBC is equal to the angle ACB; therefore (I. 4, part 2) the triangle DBC is equal to the triangle ACB, the less to the greater; which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. B Cor. Hence every equiangular triangle is also equilateral. axiom, which ought not to be done without a good reason. In other respects this proof is perfectly satisfactory; and on account of its ease and simplicity, it may perhaps be preferred by those who are reading the Elements for the first time. This proposition may also be proved by conceiving the triangle BAC to be inverted, and to be applied to the space which it before occupied, so that the point A may retain its position, while AC will fall on AB; then will the point C fall on B, AB on AC, and B on C; and the angles ACB, ECB will coincide with ABC, DBC respectively, and (I. ax. 8) be equal to them. It may be remarked that Euclid's proof, when stripped of artifice, and referred to the principle of superposition, on which it, and every demonstration which has the fourth proposition for its basis, ultimately depends, is reduced to an inversion and mutual application such as that indicated above. This will appear evident from proving the equality of the triangles ABG, ACF by superposition. * This proposition is the converse of the fifth; that is, in the language of ic, the subject of this is the predicate of the fifth, and the predicate of this the sect of the fifth. Thus, it is proved in the fifth, that if two sides of a triangle be equal, the angles opposite to them are also equal; and in the sixth, that if two angles of a triangle be equal, the sides opposite to them are equal. + This demonstration is indirect. To prove AB equal to AC, we suppose them, if possible, to be unequal. Then one of them, suppose AB, being considered the greater, we cut off a part DB, terminated at one of the equal angles, which we assume as equal, if possible, to AC; and joining CD, we prove, by means of the fourth proposition, that if the supposition were true, the triangle DBC would be PROP. VII. THEOR.*-Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. If it be possible, let there be two triangles ABC, ABD, upon the same base AB, and upon the same side of it, which have their sides CA, DA, terminated in the extremity A of the base equal to one another, and likewise their sides CB, DB, that are terminated in B. C D Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal (I. 5) to the angle ADC. But the angle ACD is greater (I. ax. 9) than BCD; therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than BCD. Again, because CB is equal to DB, the angle BDC is equal (I. 5) to BCD; but it has been demonstrated to be greater than it; therefore the supposition, that AC is equal to AD, and likewise BC to BD, is false. A C D B E P But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; therefore, because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other side of the base CD are equal (I. 5) to one another; but the angle ECD is greater (I. ax. 9) than BCD; wherefore the angle FDC is likewise greater than BCD: much more then is BDC greater than BCD. Again, because CB is equal to DB, the angle BDC is equal (I. 5) to BCD; but BDC has been proved to be greater than the same BCD; and therefore AC cannot be equal to AD, and at the same time BC to BD. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore on the same base, and on the same side of it, &c. A B equal in magnitude to ACB: but this being contrary to the 9th axiom, we conclude that AB and AC cannot be unequal, since the supposition that they are so, leads to an absurd conclusion. The student may exercise himself usefully in proving this proposition by producing CA through the vertex, and by supposing the whole line thus produced to be equal, if possible, to AB: and he may in like manner have two variations of the proof by supposing AC greater than AB. It would perhaps remove, in some degree, the difficulty which beginners often feel with respect to negative demonstrations, were they expressed hypothetically. Thus, in the present instance, we might proceed briefly in the following manner:For, if AB were not equal to AC, one of them, as AB, would be the greater. Then some part of AB, as BD, would be equal to AC; and, CD being joined, there would be two triangles DBC, ACB, in which there would be two sides and the contained angle of the one respectively equal to two sides and the contained angle of the other, and therefore (I. 4) the triangle DBC would be equal to ACB, which (I. ax. 9) is impossible. Therefore AB is not unequal to AC, that is, it is equal to it. It is often stated that this proposition affords the first instance in which the indirect method of proof is employed. This, however, is incorrect, as it is on this principle that the bases and the triangles themselves are proved to coincide in the fourth proposition, by showing that their failing to coincide would be contrary to the corollaries to the 3d and 5th definitions. * This proposition is merely a lemma to the 8th, and may be omitted, if the second method of demonstrating that proposition be adopted. PROP. VIII. THEOR.-If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal: (1.) the angle which is contained by the two sides of the one is equal to the angle contained by the two sides, equal to them, of the other; and (2.) the two triangles are equal to one another.* Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and also the base BC equal to the base EF; the angle BAC is equal to the angle EDF, and the two triangles are equal in area. B A C D G F For, if the triangle ABC be applied to DEF, so that the point B may be on E, and the straight line BC on EF; the point C shall also coincide with F, because BC is equal (hyp.) to EF. Therefore, BC coinciding with EF, BA and AC shall coincide with ED and DF; for, if the base BC coincide with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation, as EG, FG; then, upon the same base EF, and upon the same side of it, there would be two triangles having their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity; but (I. 7) this is impossible: therefore, if the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore also the angle BAC coincides with the angle EDF, and (I. ax. 8) is equal to it. Also (I. def. 5, cor.) the triangle ABC coincides with the triangle DEF, and (1. ax. 8) is equal to it. Therefore, if two triangles, &c. A OTHERWISE :— -Let the triangle ABC be inverted with respect to the base BC, and let BC be applied to EF, so that B may fall on E; then will C coincide with F, because BC is equal to EF; and the triangle ABC will take the position GEF, EG being the same as AB, and FG the same as CA. Join DG; and, because (hyp) DE and EG are equal, the angles EDG, EGD are (I. 5) equal. It would G F be shown in a similar manner, that the angles FDG, FGD are equal; and, therefore, (I. ax. 2) the angle EGF, that is, BAC, is equal to EDF. But (hyp.) the sides BA, AC are respectively equal to ED, DF, and it has now been shown that the contained angles are equal; therefore (I. 4) the triangles are equal, and the remaining angles in the one are respectively equal to the remaining angles in the other, that is, ABC to DEF, and ACB to DFE.+ * Or, if the three sides of one triangle be equal to the three sides of another, each to each; (1) the angles of the one are equal to the angles of the other, each to each, viz. those to which the equal sides are opposite; and (2) the triangles are equal to one another. † Should DG fall on DF and FG, or on DE and EG, the proof would be had by PROP. IX. PROB.-To bisect a given rectilineal angle, that is, to divide it into two equal angles. Let BAC be the given angle; it is required to bisect it. Take any point D in AB, and from AC cut off (I. 3) AE equal to AD; join DE, and upon it describe (I. 1) an equilateral triangle DEF, on the side remote from A; then join AF: AF bisects the angle BAC. Because AD is equal (const.) to AE, and AF is common to the two triangles DAF, EAF; the two sides DA, AF are equal to the two EA, AF, each to each; and the base DF is equal (const. and I. def. 17) to the base EF: therefore, the angle DAF is equal (1. 8) to EAF; wherefore, the given angle BAC is bisected by the line AF: which was to be done.t B A PROP. X. PROB.‡-To bisect a given finite straight line. Let AB be the given line: it is required to bisect it. Describe (I. 1) upon it an equilateral triangle ABC, and bisect (I. 9) the angle ACB by the straight line CD: AB is bisected in the point D.§ C Because AC is equal to CB, and CD common to the two triangles ACD, BCD; the two sides AC, CD, are equal to BC, CD, each to each; and the angle ACD is equal (const.) to the angle BCD: therefore the base AD is equal (I. 4) to the base DB, and the line AB is bisected in the point D: which was to be done. A D B means of one isosceles triangle. But should the point F fall within the isosceles triangle DEG, or E within DFG, two isosceles triangles would still be requisite, and the equality of the vertical angles would be proved by means of the third axiom. * The expression, "on the side remote from A," is added in this edition, because, if the equilateral triangle were described on the other side, its vertex might fall on the vertex of the angle to be bisected (as it would do in the next proposition), in which case the solution would fail, as no second point in the bisecting line would then be determined. In the practical construction of this problem, arcs may be described from D and E as centres with any radius greater than the half of DE, and the line joining either of their points of intersection with A will bisect the given angle. It is plain, that if the radius be taken equal to AD or AE, A must be joined with the remote point of intersection, as A and the other would coincide. If the angles BAF, CAF were bisected, the given angle would be divided into four equal parts; the bisection of which would divide it into eight equal parts; and thus the division, by successive bisections, might be continued without limit. + Otherwise.-Let DAE (1st fig. to prop. 5) be the given angle; in AD assume any points, B and F, and cut off AC and AG, respectively equal to AB and AF: join BG and CF, and the straight line joining their point of intersection with A bisects the angle. The proof, which is easy, is left to exercise the learner. This problem is a particular case of the ninth proposition of the sixth book. § In practice, the construction is effected more easily by describing ares on both sides of AB, from A as centre, and with any radius greater than the half of AB; and then, by describing arcs intersecting them, with an equal radius, from B as centre: the line joining the two points of intersection will bisect AB. The proof is easy. PROP. XI. PROB.*-To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be a given straight line, and C a point given in it; it is required to draw a straight line from C at right angles to AB. Take any point D in AC, and make (I. 3) CE equal to CD; upon DE describe (I. 1) the equilateral triangle DFE, and (I. post. 1) join FC; FC is the required line. F C E B Because DC is equal to CE, and FC common to the two triangles DCF, ECF; the two sides DC, CF are equal to the two EC, CF, AD each to each; and the base DF is equal (const. and I. def. 17) to the base EF; therefore, the angle DCF is equal (I. 8) to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called (I. def. 8) a right angle: therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB: which was to be done. PROP. XII. PROB.-To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the given straight line, which may be produced any length both ways, and let C be a point without it. It is required to draw a straight line from C perpendicular to AB. A C B E G D Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe (I. post. 3) the circle EDF meeting AB in E and F: bisect (I. 10) EF in G, and join CG; the straight line CG is the perpendicular required.‡ Join CE, CF. Then, because EG is equal to GF, and CG com * This proposition and the following contain the only two distinct cases of drawing a perpendicular to a given straight line through a given point; the first, when the point is in the line, the second, when it is without it. In the practical construction of this problem, it is sufficient to describe arcs from D and E as centres, with any radius greater than DC, and to join their point of intersection with C. The proof follows from the construction and the eighth proposition. It will be a check on the manual operation, if intersections be made on both sides of AB, as these intersections and C should all be in the same straight line, as will appear from the corollary to the 14th proposition of this book. In practice the construction will be made rather more simple, by describing from F and E, when found, arcs on the remote side of AB from C, with any radius greater than the half of FE, and joining their point of intersection with C. The 12th proposition would be better placed after the 16th; as we cannot prove, without the assistance of the 16th, that there can be but one perpendicular drawn to a straight line from a point without it. We might then infer also, that a circle can cut a straight line in only two points, since if it could cut it in more than two, there might be more than one perpendicular. |