straight line A B upon D E, the straight line A C will coincide with DF, because the angle B A C is equal to ED F. Also the point B will coincide with E, because A B is equal to DE, and the point C with F, because AC is equal to DF; and, because the points B, C, coincide with the points E, F, the straight line B C coincides with the straight line E F (ax. 10.), and (ax. 11.) is equal to it; the angle ABC coincides with DE F, and is equal to it; and the angle A C B with the angle D F E, and is equal to it. Therefore, &c, Cor. The two triangles are equal also as to surface. Scholium. It is indifferent which of the two triangles DEF be taken, although in these triangles the side D E lie in opposite directions from DF; viz. to the right of it in the one, and to the left of it in the other. The same may be observed of the next proposition, and of all cases of plane triangles, which are equal in every respect. PROP. 5. (EUc. i. 26, first part of.) If two triangles have two angles of the one equal to two angles of the other, each to each, and likewise the interjacent* sides equal; their other sides shall be equal, each to each, viz. those to which the equal angles are opposite, and the third angle of the one shall be equal to the third angle of the other. Let ABC, D E F (see the last figure) be two triangles which have the two angles ABC, ACB of the one, equal to the two angles DEF, DFE of the other, each to each, and likewise the side BC equal to the side E F: their other sides shall be equal, each to each, and the third angle BAC shall be equal to the third angle E D F. For, if the triangle A B C be applied to the triangle D E F, so that the point B may be upon E, and the straight line BC upon EF, the point C will coincide with the point F, because B C is equal to E F. Also the straight line B A will coincide in direction with ED, because the angle CBA is equal to FED, and the straight line CA with FD, because the angle B C A is equal to E FD. But, if two straight lines which cut one another, coincide with other two which cut one another, it is manifest that the points of intersection must likewise coincide. Therefore, the point A coincides with D, and the sides A B, A C, coincide with the sides DE, DF, and are Interjacent sides," i. e. sides lying between, equal to them; and the angle B A C Cor. The two triangles are equal also as to surface. PROP. 6. (EUc. i. 5 & 6.) If two sides of a triangle be equal to one another, the opposite angles shall be likewise equal: and conversely, if two angles of a triangle be equal to one another, the opposite sides shall be likewise equal. Let A B C be an isosceles triangle, having the side A B equal to the side A C; the angle A C B shall be equal to the angle A B C. Let the angle B A C be divided into two equal angles by the straight line AD, which meets the base BC in D (Post. 4). Then, because the triangles AD B, ADC have two sides of the one equal to two sides of the other, each to each, and the interjacent angles BAD, CAD equal to one another, their other angles are equal, each to each (4.); therefore the angle ACB is equal to ABC. Next, let the angle ABC be equal to the angle ACB: the side A C shall be equal to the side A B. From D, the middle point of BC, erect a perpendicular to B C (Post. 3. and 5.): and, if it do not pass through the vertex A, let this perpendicular, if possible, cut one of the sides as A B in E, and join E C. Then, because the triangles ED B, EDC have two sides of the one equal to two sides of the other, each to each, and the included angles ED B, EDC equal to one another (def. 10.), their other angles are equal, each to each (4.). Therefore the angle ECD is equal to EBD. But EBD or ABD is equal to ACB: therefore the angle ECD is equal to ACB (ax. 1.), the less to the greater, which is impossible. Therefore the perpendicular at D cannot pass otherwise than through the vertex A: and because the triangles AD B, ADC are equal, according to Prop. 4., the side A B is equal to the side A ̊C. Therefore, &c. Cor. 1. Every equilateral triangle is also equiangular; and conversely. Cor. 2. In an isosceles triangle ABC, if the equal sides A B, A C, be produced, the angles upon the other side of the base BC will be equal to one another; for, each of them together with one of the equal angles A B C, A CB, is equal to two right angles (2.). Cor. 3. The straight line which bisects the vertical angle of an isosceles triangle, bisects the base at right angles: and conversely, the straight line which bisects the base at right angles, passes through the vertex, and bisects the vertical angle. Cor. 4. If there be two isosceles triangles upon the same base (whether they be upon the same side of it or upon different sides), the straight line which joins their vertices or summits, or that straight line produced, shall bisect the base at right angles. For the straight line which bisects the base at right angles passes through the vertex of each. PROP. 7. (Euc. i. 8.) 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, the angle contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them of the other. Let ABC, DEF be two triangles, having the two sides of the one equal to two sides of the other, each to each, and likewise the base BC equal to the base EF: the angle BAC shall be equal to the angle ED F. At the point E in the straight line EF, make the angle FEG equal to the angle A B C (Post. 6.): take EG equal to BA or E D, and join G F, G D. Then because the triangles A B C, GEF, have two sides of the one equal to two sides of the other, each to each, and the included angles equal to one another, the base G F is equal to A C (4.) that is, to DF. Again, because in the triangle FDG, the side FD is equal to FG, the angle FDG is equal to FGD (6.). For the like reason, the angle E D G is equal B B to EGD. Therefore the angle EDF, which is the sum or difference of the two EDG, FD G, is equal to the angle EG F, which is the sum or difference of the two E GD, FGD (ax. 2., 3.). But EGF is equal to B A C, because (4.) the triangle GEF is equal to the triangle ABC in every respect: therefore, (ax. 1.) the angle EDF is equal to the angle BAC. When G D coincides with GE, GED is a straight line, and the angles at G and D are the angles at the base of the isosceles triangle FDG; wherefore the latter is equal to the angle at G, that is, to the angle at A, as before. Therefore, &c. Cor. The two triangles are equal in every respect. (4. & 4 Cor.) PROP. 8. (EUc. i. 17.) Any two angles of a triangle are together less than two right angles. Let A B C be any triangle: any two of its angles, ABC and ACB, shall be together less than two right angles. Bisect B C in D (Post. 3.): join A D, and produce it to E, so that DE may be equal to A D; and join CE. Then, because the triangles ABD ECD have two sides of the one equal to the two sides of the other, each to each, and the included angles A D B, E DC, equal to one another (3.), the angle ECD is equal to the angle ABD or AB C. (4.) Therefore, the two angles A B C, AC B taken together are equal to the two angles ECD, ACB taken together (ax. 2.), that is, to the angle ACE. But (2. Cor. 2.) the angle ACE is less than two right angles. Therefore, the angles ABC, ACB together are less than two right angles. Therefore, &c. Cor. 1. (Euc. i. 16.) If one side of a triangle ABC, as BC, be produced to F, the exterior angle A C F shall be greater than either of the interior and opposite angles at A and B; for either of these angles taken with the angle ACB, is less than two right angles, but the angle A CF, taken with the same ACB, is equal to two right angles (2.). Cor. 2. A triangle cannot have more than one right angle, or more than one obtuse angle. PROP. 9. (EUC. i. 18. & 19.) If one side of a triangle be greater than another, the opposite angle shall B Take A D equal to A C, and join CD. Then, because AD is equal to A C, the angle A CD is equal to the angle ADC (6.). But, because the side B D of the triangle C D B is produced to A, the exterior angle ADC is greater than the interior and opposite angle D B Cor A B C. (8.Cor. 1.) Therefore, the angle ACD, and much more AC B, is also greater than ABC. Next, let the angle A C B be greater than the angle ABC; the side AB shall likewise be greater than the side AC. For, AB cannot be equal to AC; because, then, (6.) the angle ACB would be equal to A B C, which is not the case: neither can it be less than AC, because then, by the former part of the proposition, the angle ACB would be less than A B C, which is not the case. Therefore AB cannot but PROP. 10. (Euc. i. 20.) Any two sides of a triangle are together greater than the third side: and any side of a triangle is greater than the difference of the other two. Let A B C be a triangle: any two of its sides, A B and A C, shall be together greater than the third side BC; and any side A B alone shall be greater than the difference of BC and A C, the other two sides. R gether greater than BC, if A C be taken Л Cor. 1. (Euc. i. 21, part of.) If there be two triangles ABC, DBC, upon the same base BC, and if the vertex of one of them, as D, fall within the other, the B two sides of that triangle will be less than the two sides of the other. For, if CD be produced to meet the side A B of the enveloping triangle in E, BD and DC together will be less than B E and EC together, (ax. 6.) because B D is less than BE and ED together: and, for the like than BA and AC together: much reason, BE and EC together are less more, then, are B D and DC together less than BA and A C together. is less than the sum of all the other sides. Cor. 2. Any side of a rectilineal figure Cor. 3. And, hence, it may easily be demonstrated, that if there be two rectilineal figures ADC, D B C upon the same base BC, one of which other, the perimeter of wholly envelopes the the enveloping figure must be greater than the perimeter of the other. Scholium. B C By help of this proposition it may be shown that a straight line is the shortest distance between two points A and B. Let A CB be the straight line joining A and B, and ADEB any other line drawn from A to B. In ACB take any point C; and from the centre B in D; and join AD, DB. Then, because A D and D B are together greater Produce B A to D, so that A D may than A B, and that A D is equal to A C, be equal to A C, and join C D. Then, DB is greater than CB (ax. 6.). Therebecause A D is equal to A C, the angle fore, if a circle be described from the ACD is equal to A D C (6.) But the centre B with the radius B C, it will cut angle BCD is greater than ACD: the straight line D B in some point betherefore, the angle B CD is greater tween D and B; and, consequently, the also than ADC or B D C. Therefore, line ADEB in some point E which is (9.) the side BD is, likewise, greater in the part DE B. Join E B. Then, than B C. But, BD is equal to B A if AD be made to coincide with A C, and A C together, because A D is equal and BE with B C, it is evident that the to A C. Therefore BA and AC to- parts AD and E B (curvilineal or gether are greater than B C. otherwise) of the whole line ADEB And, because BA and AC are to- (curvilineal or otherwise) will form a complete path from A to B, which is shorter than ADEB by the intermediate part D E. Therefore, there is no path from A to B, the straight line A C B excepted, than which a shorter may not be found between A and B. But, since none of the paths from A to B can be less than of some certain length, there must be some, one or more, shorter than the others. Therefore, the shortest path is the straight line AC B. From this property the straight line which joins two points derives the name of the distance between them. C Hence, also, we may infer, that of any two paths, ACB, AD B, leading from A to B, and everywhere concave towards the straight line AB, that which is enveloped by the other, as ADB, is the shortest. For of all the paths not lying between ADB and the straight line A B, there is none, A DB excepted, than which a shorter may not be found. And this is the case whether the paths ACB and ADB be both of them curvilineal, or one of them, (ACB or ADB) rectilineal. PROP. 11. (Euc. i. 24 & 25.) If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle which is contained by the two sides of the one greater than the angle which is contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other: and conversely. Let А В С, DE F, be two triangles having the two sides AB, AC of the one equal to the two sides DE, DF of the other, each to each, but the angle B A C greater than the angle EDF: the base, B C shall be greater than the base E F. At the point D in the straight line DE, make the angle EDG equal to the angle BAC (Post. 6.); take DG equal to AC, and join EG, GF. Then, because the triangles ABC, DEG have two sides of the one equal to two sides of the other, each to each, and the included angles BAC, EDG equal to one another, the bases BC, EG are equal to one another (4.). Now, the line D F falls between DE and DG, because the angle EDG is equal to BAC, which is supposed to be greater than ED F. But the point F may fall 1° without the triangle DEG; or 20 upon the base EG; or 3° within the triangle D E G. In the first case, because DG is equal to A C, that is, to D F, the angle DFG is equal to the angle DG F (6.). But the angle EFG is greater than DFG, and EGF is less than DG F. Therefore much more is the angle E FG greater than the angle EGF. Therefore, also, the side E G, that is B C, is greater than E F (9.). In the second case, it is at once evident that EG, that is, B C, is greater than E F. D AA B 0 E F one another (6. Cor. 2.). But EFG is greater than H FG, and EGF is less than KG F. Therefore much more is the angle EFG greater than the angle EGF, as in the first case, and the side EG, that is B C, is greater than E F (9.). Next, let the base B C be greater than the base EF; the angle BAC shall likewise be greater than the angle EDF. For B A C cannot be equal to EDF, because then (4.) the base BC would be equal to the base EF; neither can it be less than E D F, because then, by the former part of the proposition, the base BC would be less than the base EF. Therefore, the angle B A C cannot but be greater than the angle EDF. Therefore, &c. PROP. 12. A straight line may be drawn perpendicular to a given straight line of indefinite length, from any given point without it; but, from the same point, there cannot be drawn more than one perpendicular to the same straight line. Let BC be a given straight line of indefinite length, and A any given point without it. A perpendicular may be drawn from the point A to the straight line B C. T E In BC take any point D; join AD, and produce it to any point E, (Post. 1.). With the centre A and the radius A E describe a circle cutting B C in the points B and C upon each side of the point D, (Post. 2.). Bisect BC in F, and join AF, AB, AC, (Post. 3.). Then because AB is equal to A C, ABC is an isosceles triangle. Therefore AF, which is drawn from the vertex A to the middle point of the base BC, is perpendicular to the base (6. Cor. 3.); that is, a straight line AF may be drawn from the point A perpendicular to the straight line B C. But, from the same point A there cannot be drawn more than one perpendicular to the same straight line B C. For, if any other straight line AD were perpendicular to B C, the two angles, ADF and AFD, of the triangle AD F, would be together equal to two right angles, which (8.) is impossible. Therefore, &c. Cor. 1. If from any point A to a straight line B C, there be drawn a straight line AB which is not at right angles to B C, a second straight line AC may be drawn from A to B C, which shall be equal to AB; for, the perpendicular AF being drawn, and FC being taken equal to FB, it may easily be shown (4.) that A C is equal to A B. Cor. 2. Of straight lines A B, AD, which are drawn from A to B C upon the same side of the perpendicular Á F, that which is nearer to the perpendicular, as AD, is less than the other, which is more remote. For, the angle A B D or ABF being (8. Cor. 1.) less than the exterior right angle AFC, and again AFD or AFC less than the exterior angle AD B, much more is the angle ABD less than AD B, and therefore also the side AD less than AB (9.). Cor. 3. In the same manner it may be shown that the perpendicular A F is the least of all straight lines which can be drawn from A to B C. For, if A B be any other straight line, the angle ABF being less than the exterior angle AFC, that is than AFB (def. 10.), the side A F is also less than the side A B (9.). For this reason, the perpendicular AF is called also the distance of the point A from the line B C. Cor. 4. Hence, if, from the centre A, a circle be described with a radius less than the perpendicular A F, it will not meet the straight line BC; if with a radius equal to A F, it will meet B C in one point only, which is the foot of the perpendicular; and if with a radius greater than A F, it will meet B C in two points, which are at equal distances from the foot of the perpendicular, upon either side of it. and side A B of the one be equal to the hypotenuse DF and side DE of the other. The triangles ABC, DEF, shall be equal to one another in every respect. For, if the side A B be made to coincide with DE, which is equal to it, the right angle A B C will also coincide with the right angle DEF (1. and ax. 11). Therefore, if AC do not coincide with DF, but fall otherwise, as DG, there will be drawn from the point D to the line E F, upon the same side of the perpendicular, two straight lines that are equal to one another, which is impossible (12. Cor. 2.). Therefore, A C coincides with DF, and the triangle A B C coincides with the triangle DEF, that is, (ax. 11.) the triangles ABC and DEF are equal in every respect. Next, let the hypotenuse AC and angle ACB of the one triangle be equal to the hypotenuse D F and angle DFE of the other. In this case, also, the triangles shall be equal in every respect. For if the hypotenuse A C be made to coincide with D F, which is equal to it, the angle ACB will also coincide with DFE, which is equal to it. Therefore, if A B do not coincide with DE, but fall otherwise, as DG there will be |