Book I. ADB is equal to the angle CBD; and by Prop. 4. B. 1, the rangle BAD is equal to the angle DCB, and ABD to BDC; and therefore also the angle ADC is equal to the angle ABC. And if the angle BAD be equal to the opposite angle BCD, and the angle ABC to ADC; the opposite sides are equal : Because, by Prop. 32. B. 1, all the angles of the quadrilateral figure ABCD are together equal to four right angles, and the two angles A BAD, ADC are together equal to the two angles BCD, ABC: Wherefore BAD, ADC are the balf of all B the four angles; that is, BAD and ADC are equal to two right angles : And therefore AB, CD are parallels by Prop. 28. B. 1. In the same manner, AD, BC are parallels. Therefore ABCD is a parallelogram, and its opposite sides are equal by the 34th Prop. B. 1. PROP. VII. B. I. . There are two cases of this proposition, one of which is not in the Greek text, but is as necessary as the other: And that the case left out has been formerly in the text, appears plainly from this, that the second part of Prop. 5, which is necessary to the demonstration of this case, can be of no use at all in the elements, or any where else, but in this demonstration; because the second part of Prop. 5. clearly follows from the first part, and Prop. 13. B. 1. This part must therefore have been added to Prop. 5, upon account of some proposition betwixt the 5th and 13th, but none of these stand in need of it except the 7th proposition, on account of which it has been added : Besides, the translation from the Arabic has this case explicitly demonstrated. And Proclus acknowledges, that the second part of Prop. 5. was added upon account of Prop. 7, but gives a ridiculous reason for it, “ that it might 6 afford an answer to objections made against the 7th,” as if the case of the 7th, which is left out, were, as he expressly makes it, an objection against the proposition itself. Whoever is curious may read what Proclus says of this in his commentary on the 5th and 7th Propositions; for it is not worth while to relate his trifies at full length. It was thought proper to change the enunciation of this 7th Prop. so as to preserve the very same meaning ; the literal translation from the Greek being extremely harsh and difficult to be understood by beginners, PROP. XI. B. I. Book I. A Corollary is added to this proposition, which is necessary to Prop I. 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 in-, cluded 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 DM FG , G H FH, betwixt F and H, because FD is less than FH; and that, for the like reason, the circle described from the centre G, at the distance GH or GM, must meet DG betwixt D and G; Book I. and that these circles must meet one another, because FD and GH are together greater than FG? And this determina- DM FG H PROP. XXIV. B. I. D Mr Thomas Simpson, in p. 262 of 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 C B go farther from another straight D E F H For example, the straight line ABC cannot first come nearer to the straight line DE, as from the point X to the point B, and A B C then, from the point B to the point D gure above. farther from the same DE: E And, 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. See the fi. C, go 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 from AB at the point D than at F, that is, FD goes farther from AB from F to D: Therefore the straight A : E E line CFD first comes nearer to the straight line AB, and then goes farther 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 farther 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. a 4. 1. If two equal straight lines, AC, BD lie 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 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, and the base AD is equal to the base BC. Therefore the angle F D b 8. 1. ACD is equal to the angle BDC: G From any point Ein AB draw EF to CDat right angles to AB: therefore, by Prop. 1, EF is equal to AC, or BD; wherefore, as has been A Е В E 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. 1. 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 B a |