PROP. XI. B. I. A corollary is added to this proposition, which is necessary to prop. 1. B. 11. and otherwise. PROP. XX. and XXI. B. I. Proclus, in his commentary, relates, that the Epicureans derided this proposition, as being manifest even to asses, and needing no demonstration; and his answer is, that though the truth. of it be manifest to our senses, yet it is science which must give the reason why two sides of a triangle are greater than the third: but the right answer to this objection against this and the 21st. and some other plain propositions, is, that the number of axioms ought not to be increased without necessity, as it must be if these propositions be not demonstrated. Mons. Clairault, in the preface to his Elements of Geometry, published in French at Paris anno 1741, says, That Euclid has been at the pains to prove, that the two sides of a triangle which is included within another are together less than the two sides of the triangle which includes it; but he has forgot to add this condition, viz. that the triangles must be upon the same base: because, unless this be added, the sides of the included triangle may be greater than the sides of the triangle which includes it, in any ratio which is less than that of two to one, as Pappus Alexandrinus has demonstrated, in prop. 3. B. 3. of his mathematical collections. PROP. XXII. B. I. Some authors blame Euclid, because he does not demonstrate that the two circles made use of in the construction of this problem must cut one another: but this is very plain from the determination he has given, viz. that any two of the straight lines DF, FG, GH must be greater than the third: for who is so dull, though only beginning to learn the Elements, as not to perceive that the circle described from the centre F, at the distance FD, must meet FH betwixt F and H, because FD is less than FH; D M FG H and that, for the like reason, the circle described from the centre G, at the distance GH, or GM, must meet DG betwixt D DM FG H and G; and that these circles must meet one another, because FD and GH are together greater than FG? and this determination is easier to be understood than that which Mr. Thomas Simson derives from it, and puts instead of Euclid's in the 49th page of his Elements of Geometry, that he may supply the omission he blames Euclid for; which determination is that any of the three straight lines must be less than the sum, but greater than the difference of the other two: from this he shows the circles must meet one another, in one case; and says that it may be proved after the same manner in any other case; but the straight line GM, which he bids take from GF, may greater than it, as in the figure here annexed: in which case his demonstration must be changed into another. PROP. XXIV. B. 1. To this is added, "of the two sides DE, DF, let DE be that which is not greater than the other;" that is, take that side of the two DE, DF, which is not greater than the other, in order to make with it the angle EDG equal to BAC, because, without this restriction, there might be three different cases of the proposition, as Campanus and others make. D & Mr. Thomas Simson, in p. 262 of the second edition of his Elements of Geometry, printed anno 1760, observes, in his notes, that it ought to have been shown that the point F falls below the line EG. This probably Euclid omitted, as it is very easy to perceive, that E DG being equal to DF, the point G is in the circumference of a circle described from the centre D, at the dis tance DF, and must be in that part of it which is above the straight line EF, because DG falls above DF, the angle EDG being greater than the angle EDF. PROP. XXIX. B. I. The proposition which is usually called the 5th postulate, or 11th axiom, by some the 12th, on which this 29th depends, has given a great deal to do, both to ancient and modern geometers: it seems not to be properly placed among the axioms; as indeed it is not self-evident; but it may be demonstrated thus: DEFINITION 1. The distance of a point from a straight line, is the perpendicular drawn to it from the point. DEF. 2. One straight line is said to go nearer to, or further from another straight line, when the distance of the points of the first from the other straight line becomes less or greater than they were; and two straight lines are said to keep the same distance from one another, when the distance of the points of one of them from the other is always the same. AXIOM. A straight line cannot first come nearer to another straight line, and then go further from it, before it A cuts it; and, in like manner, a straight D. B C E H distance from another straight line, and then come nearer to it, or go further from it; for a straight line keeps always the same direction. A B For example, the straight line ABC cannot first come nearer to the straight line DE, as from the point A to the point B, and then, D from the point B to the point C, go F further from the same DE: and, in like manner, the straight line FGH G C E H cannot go further from DE, as from F to G, and then, from G to H, come nearer to the same DE: and so in the last case, as in figure 2. (See the figure above.) PROP. I. If two equal straight lines, AC, BD, be each at right angles to the same straight line AB; if the points C, D be joined by the straight line CD, the straight line EF drawn from any point E in AB unto CD, at right angles to AB, shall be equal to AC, or BD. If EF be not equal to AC, one of them must be greater than the other; let AC be the greater; then, because FE is F D F less than CA, the straight line CFD is nearer to the straight line AB at the point F than at the point C, that is, CF comes nearer to AB from the point C to F: but because DB is greater than FE, the straight line CFD C is further from AB at the point D than at F, that is, FD goes further from AB from F to D: therefore the straight line CFD first comes nearer to the A straight line AB, and then goes further from it, before it cuts it; which is impossible. If FE be said to be greater than CA, or DB, the straight line CFD first goes further from the straight line AB, and then comes nearer to it; which is also impossible. Therefore FE is not unequal to AC, that is, it is equal to it. PROP. II. E B If two equal straight lines AC, BD be each at right angles to the same straight line AB; the straight line CD, which joins their extremities, makes right angles with AC and BD. F Ꭰ Join AD, BC; and because, in the triangles CAB, DBA, CA, AB are equal to DB, BA, and the angle CAB equal to the angle DBA; the base BC is equal (4. 1.) to the base AD: and in the triangles ACD, BDC, AC, CD, are equal to BD, DC, and the base AD is equal to the base BC: C therefore the angle ACD is equal (8. 1.) to the angle BDC from any point E in AB draw EF unto CD, at right angles to AB: therefore by prop. 1. EF is equal to AC, or BD; wherefore, A E B G as has been just now shown, the angle ACF is equal to the angle EFC in the same manner, the angle BDF is equal to the angle EFD; but the angles ACD, BDC are equal; therefore the angles EFC and EFD are equal, and right angles (10. def. 1.); wherefore also the angles ACD, BDC are right angles. COR. Hence, if two straight lines AB, CD be at right angles to the same straight line AC, and if betwixt them a straight line BD be drawn at right angles to either of them, as to AB; then BD is equal to AC, and BDC is a right angle. If AC be not equal to BD, take BG equal to AC, and join CG; therefore, by this proposition, the angle ACG is a right angle; but ACD is also a right angle; wherefore the an gles ACD, ACG are equal to one another, which is impossible. Therefore BD is equal to AC; and by this proposition BDC is a right angle. PROP. III. If two straight lines which contain an angle be produced, there may be found in either of them a point from which the perpendicular drawn to the other shall be greater than any given straight line. Let AB, AC be two straight lines which make an angle with one another, and let AD be the given straight line; a point may be found either in AB or AC, as in AC, from which the perpendicular drawn to the other AB shall be greater than AD. F K B M In AC take any point E, and draw EF perpendicular to AB; produce AE to G, so that EG be equal to AE, and produce FE to H, and make EH equal to FE, and join HG. Because, in the triangles AEF, GEH, AE, EF are equal to GE, EH, each to each, and contain equal (15. 1.) angles, the angle GHE is therefore equal (4. 1.) to the angle AFE, which is a right angle: draw GK perpendicular to AB; and because the straight lines FK, HG are at right angles to FH, and KG at right angles to FK; N KG is equal to FH, by cor. pr. 2. that is, to the double of FE. In the same manner, if AG be produced to L, so that D A G C L GL be equal to AG, and LM be drawn perpendicular to AB, then LM is double of GK, and so on. In AD take AN equal to FE, and AO equal to KG, that is, to the double of FE, or AN; also take AP equal to LM, that is, to the double of KG, or AO; and, let this be done till the straight line taken be greater than AD; let this straight line so taken be AP, and because AP is equal to LM, therefore LM is greater than AD. Which was to be done. PROP. IV. If two straight lines AB, CD make equal angles EAB, ECD with another straight line EAC towards the same parts of it; AB and CD are at right angles to some straight line. |