Book I. and that these circles must meet one another, because FD and GH are together greater than FG? And this determina- PROP. XXIV. B. I. D Mr Thomas Simpson, in p. 262 of G falls below the line EG. This pro H bably Euclid omitted, as it is very easy to perceive, that DG being equal to DF, the point G is in the circumference of a circle described from the centre D at the distance 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 geo- Book I. meters: It seems not to be properly placed among the axioms, as indeed it is not self evident; but it may be demonstrated thus : DEFINITION I. 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 near to, or farther from, another straight line, when the distances of the points of the first from the other straight line become 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 farther from it, before it cuts it; and, in like A manner, a straight line cannot B C go farther from another straight D line, and then come nearer to it ; nor can a straight line keep F G H the same distance from another straight line, and then come nearer to it or go farther from it; for a straight line keeps always the same direction. 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 A. B See the fi С then, from the point B to the point D C, go farther from the same DE: E Anå, in like manner, the straight F G H line FGH cannot go farther 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 fig. 2. 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 to 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 gure above. T Book I. less than CA, the straight line CFD is nearer to the straight line AB at the point F than at the F D F A E B PROP. II. b 8. 1. 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. Join AD, BC; and because, in the triangles CAB, DBA, CA, AB are equal to DB, BA, and the angle CAB equal to a 4. 1. the angle DBA; the base BC is equal a to the base AD: And in the triangles ACD, BDC, AC, CD are equal to BD, DC, F D G From any point E in A B draw EF to CDat right angles to AB: therefore, by Prop. 1, EF is equal to AC, or BD; wherefore, as has been A E B 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, c 10. def. I. and right angles , 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 HG. right angle; but ACD is also a right angle: wherefore the Book 1. angles ACD, ACG are equal to one another, which is impossible. Therefore B.D 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. In AC take any point E, and draw EF perpendicular to AB; produce AE to G, so that EG may be equal to AE; and produce FE to H, and make EH equal to FE, and join Because, in the triangles A EF, GEH, AE, EF are equal to GE, EH, each to each, and contain equal a angles, a 15. 1. the angle GHE is therefore equal to the angle AFE which b 4. 1. is a right angle: Draw GK perpendicular to AB; and because the straight lines FK, HG are at А. F K B M right angles to FH, and KG at NI right angles too! E FK, KG is equal H' to FH, by Cor. С Pr. 2, that is, to P! L PROP. IV. G Book I. b 4. 1. Bisect AC in F, and draw FG perpendicular to AB; take CH in the straight line CD equal to AG, and on the contrary side of AC to that on which AG is, and join FH: Therefore in the triangles AFG, CFH, the sides FA, AG are equal to FC, CH, each to each, and the angle E GA B HFA, which two last are equal to- also AFG, AFH are equal to two d 14. 1. right angles, and consequently d GF and FH are in one straight line. And because AGF is a right angle, CHF, which is equal to it, is also a right angle: Therefore the straight lines AB, CD are at right angles to GH. PROP. V. If two straight lines AB, CD, be cut by a third ACE, so as to make the interior angles BAC, ACD, on the same side of it together less than twv right angles ; AB and CD being produced shall meet one another towards the parts on which are the two angles which are less than two right angles. At the point C, in the straight line CE, make a the angle ECF equal to the angle EAB, and draw to A B the straight line CG at right angles to CF: Then, because the angles ECF, EAB are equal to one another, and the E angles ECF, FCA, are b 13. 1. together equal to two right angles, the angles МС, K EAB, FCA are equal two right angles. N D L H fore the angle FCA is greater than ACD, and CD falls between CF and AB: And because CF and CD make an angle with one another, by Prop. 3, a point may be found in either of them CD, from which the perpendicular drawn to CF shall be greater than the straight line CG. Let this point be H, and draw a 23. 1. F to |