PROP. XVI. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior remote angles. Let ABC be a triangle, and let its side BC be produced to D; the exterior angle ACD is greater than either of the interior remote angles CBA, BAC. Bisect (I. 10.) AC in E, join (I. post. 1.) BE, and produce it (I. post. 2.) to F; make (I. 3.) EF equal to BE; and join FC. B A F E -D HC Because (const.) AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angles AEB, CEF are equal (I. 15.) because they are opposite vertical angles; therefore the angle BAE is equal (I. 4. part 3.) to ECF; but the angle ECD is greater (I. ax. 9.) than ECF; therefore the angle ACD is greater than BAE. In the same manner, if the side AC be produced (I. post. 2.) to G, and BC be bisected in H, and if a straight line be drawn from A to H and produced till the produced part be equal to AH, and if the extremity of that part be joined with C, it may be demonstrated that the angle BCG, that is, (I. 15.) ACD, is greater than ABC. Therefore, if one side, &c. Cor. 1. Hence from a point without a straight line, only one perpendicular can be drawn to the line. For, if there could be two, a triangle would be formed which would have an exterior angle equal to an interior and remote one, each being a right angle; which, by the proposition, is impossible. Cor. 2. If from a point without a line, two straight lines be drawn cutting it, one of them at right angles to it, and the other not, the perpendicular will fall on that side of the other on which the acute angle is: for if it fell on the other, the exterior acute angle would be greater than the interior and remote right angle, which (I. def. 10.) is impossible. PROP. XVII. THEOR.* ANY two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles are together less than two right angles. ing the equality of CEB and AED, either AEC or BED may be made the common angle. It is also evident that when AEC and BED have been proved to be equal, the equality of AED and BEC, might be inferred from the 13th proposition and the third axiom. * This proposition might be omitted, as it is included in the 32d of this book, and is not referred to in any thing preceding that proposition. B A C D Produce BC to D; and because ACD is the exterior angle of the triangle ABC, it is greater than the interior and remote angle ABC; to each of these add ACB: therefore the angles ACD, ACB are greater (I. ax. 4.) than ABC, ACB; but ACD, ACB are together equal (I. 13.) to two right angles; therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB are less than two right angles; and if CB be produced through B, the same may be demonstrated respecting the angles A and B. Therefore, any two angles, &c. Cor. Hence every triangle must have at least two acute angles. PROP. XVIII. THEOR.* Ir two sides of a triangle be unequal, the greater side has the greater angle opposite to it. Let ABC be a triangle, of which the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA. B D C Because AC is greater than AB, make (I. 3.) AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater (1. 16.) than the interior and remote angle C; but (I. 5.) ADB is equal to ABD, because the sides AB, AD are equal; therefore the angle ABD is likewise greater than C; wherefore much more is the angle ABC greater than C. Therefore, if two sides, &c. PROP. XIX. THEOR. IF two angles of a triangle be unequal, the greater angle has the greater side opposite to it. Let ABC be a triangle, of which the angle B is greater than C; the side AC is likewise greater than AB. For if it be not greater, AC must either be equal to AB, or less than it. It is not equal, because then (I. 5.) the angle B Let the learner compare this proposition and the following with the 5th and 6th of this book. This proposition might be proved with equal facility by producing AB through B, till the whole line thus produced is equal to AC, and joining C with the extremity of the produced part. The following proof is very neat and easy: From AC cut off AD equal to AB; bisect (1. 9.) the angle BAC by the straight line AE, and join ED. Then in the triangles BAE, DAE, BA is equal to DA, AE common, and the angle BAE equal to DAE: therefore (I. 4. part. 3.) the angle ADE is equal to B. But (I. 16.) ADE is greater than C; therefore also B is greater than C. B E D Neither is it less; would be equal to the angle C; but it is not. PROP. XX. THEOR. B A ANY two sides of a triangle are together greater than the third side.* Let ABC be a triangle; any two sides of it are together greater than the third side. Produce (I. post. 2.) BA to the point D, and make (I. 3.) AD equal to AC; and join DC. B A D Because DA is equal to AC, the angle ADC is likewise equal (I. 5.) to ACD; but the angle BCD is greater (I. ax. 9.) than ACD; therefore the angle BCD is greater than ADC: and because in the triangle DBC, the angle BCD is greater than BDC, and that (I. 19.) the greater side is opposite to the greater angle; therefore the side DB is greater than the side BC; but DB is equal to BA and AC; therefore BA, AC are greater than BC. In the same manner, it might be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB. Therefore, any two sides, &c. Cor. Hence the difference of any two sides of a triangle, is less than the third side. For, since BA and AC are together greater than BC, from each of these take AC; then (I. ax. 5.) BA is greater than the excess of BC above AC. PROP. XXI. THEOR. + If from a point within a triangle, two straight lines be drawn to the extremities of one of the sides, these lines are together less than the other sides, but contain a greater angle. From the point D within the triangle BAC, let the straight *The truth of this proposition is so manifest, that a proof of it is given merely to avoid increasing the number of axioms. If, as is done by Archimedes, a straight line were defined to be the shortest distance between two points, this proposition would follow immediately as a corollary. A proof as easy as the one given above would be obtained by bisecting the angle BAC; as it follows from the 16th proposition, that each of the angles made by the bisecting line with BC is greater than the half of BAC; and therefore (I. 19.) each segment of BC is less than the adjacent side. This proposition is never referred to by Euclid, except in the 8th proposition of the third book; and that proposition may be proved without it. By means of the 32d of this book, it may be proved that the angle BDC exceeds BAC by the sum of the angles ABD, ACD. lines DB, DC be drawn to the extremities of BC; DB, DC are less than AB, AC, but the angle BDC, is greater than BAC. Produce (I. post. 2.) BD to E; and because two sides of a triangle are greater (I. 20.) than the third, the two sides BA, AE, are greater than BE. B DE C To each of these add EC; therefore (I. ax. 4.) BA, AC are greater than BE, EC. Again, because the two sides CE, ED are greater than CD, add DB to each; therefore CE, EB are greater than CD, DB. But it has been shown that BA, AC are greater than BE, EC: much more then are BA, AC greater than BD, DC. Again, the exterior angle BDC of the triangle CDE is greater (1. 16.) than CED; for the same reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than CEB; much more then is the angle BDC greater than BAC. Therefore, if from a point, &c. PROP. XXII. PROB. To describe a triangle of which the sides shall be equal to three given straight lines; but any two of these must be greater (I. 20.) than the third. Let A, B, C be three given straight lines, of which any two are greater than the third: it is required to make a triangle of which the sides shall be equal to A, B, C, each to each. D G H K M A B C Take an unlimited straight line DE, and let F be a point in it, and make (I. 3.) FG equal to A, FH to B, and HK to C. From the centre F, at the distance FG, describe (I. post. 3.) the circle GLM, and from the centre H, at the distance HK, describe the circle KLM. Now, because (hyp.) FK is greater than FG, the circumference of the circle GLM, will cut FE between F and K, and therefore the circle KLM cannot lie wholly within the circle GLM. cause (hyp.) GH is greater than HK, the circle GLM cannot lie wholly within the circle KLM. Neither can the circles be wholly without each other, since (hyp.) GF and HK are together greater than FH. The circles must therefore intersect each other: let them intersect in the point L, and join LF, LH; the triangle LFH has its sides equal respectively to the three lines A, B, C. In like manner, be Because F is the centre of the circle GLM, FL is equal (I. def. 30.) to FG; but (const.) FG is equal to A; therefore (I. ax. 1.) FL is equal to A. In like manner, it may be shown that HL is equal to C; and (const.) FH is equal to B: therefore the three straight lines LF, FH, HL are respectively equal to the three A, B, C and therefore the triangle LFH, has been constructed, having its three sides equal to the three given lines, A, B, C: which was to be done. * PROP. XXIII. PROB. AT a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. C Let AB be the given straight line, A the given point in it, and C the given angle; it is required to make an angle at A, in the straight line AB, that shall be equal to C. E F B In the lines containing the angle C, take any points D, E, and join them, and make D (I. 22.) the triangle AFG, the sides of which AF, AG, FG, shall be equal to the three straight lines CD, CE, DE, each to each. Then, because DC, CE, are equal to FA, AG, each to each, and the base FG to the base DE, the angle A is equal (1.8.) to the angle C. Therefore, at the given point A in the given straight line AB, the angle A is made equal to the given angle C: which was to be done. † PROP. XXIV. THEOR. IF two triangles have two sides of the one equal to two sides of the other, each to each, but the angles contained by those sides unequal: the base of that which has the greater angle is greater than the base of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each; but the angle BAC greater than EDF; the base BC is also greater than the base EF. Of the two sides DE, DF, let DE be that which is not It is evident that if MF, MH were joined, another triangle would be formed having its sides equal to A, B, C. It is also obvious that in the practical construction of this problem, it is only necessary to take with the compasses FH equal to B, and then, the compasses being opened successively to the lengths of A and C, to describe circles or arcs from F and H as centres, intersecting in L; and lastly to join LF, LH. The construction in the text is made somewhat different from that given in Simson's Euclid, with a view to obviate objections arising from the application of this proposition in the one that follows it. + The construction of this proposition is most easily effected in practice, by making the triangles isosceles. In doing this, circular arcs are described with equal radii from C and A as centres, and their chords are made equal. It is evident that another angle might be made at A, on the other side of AB, equal to C. |