definitions are distinguished from the rest by inverted double com



The words, " which also divide the circle into two equal parts,” are added at the end of this definition in all the copies, but are now left out as not belonging to the definition, being only a corollary from it. Proclus demonstrates it by conceiving one of the parts into which the diameter divides the circle, to be applied to the other ; for it is plain they must coincide, else the straight lines from the centre to the circumference would not be all equal: the same thing is easily deduced from the 31st prop. of book 3, and the 24th of the same; from the first of which it follows, that semicircles are similar segments of a circle: and from the other, that they are equal to one another.


This definition has one condition more than is necessary; because every quadrilateral figure which has its opposite sides equal to one another, has likewise its opposite angles equal, and on the contrary.

Let ABCD be a quadrilateral figure, of which the opposite sides AB, CD are equal to one another: as also AD A

D and BC: join BD; the two sides AD, DB are equal to the two CB, BD, and the base AB is equal to the base CD, therefore, by prop. 8, of book 1, the angle ADB is equal to the angle


С CBD; and, by prop. 4, book 1, the angle 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, book 1, all the angles of the quadrilateral figure ABCD are together equal to four right angles, and the A

D two angles BAD, ADC, are together equal to the two angles BCD, ABC: wherefore BAD, ADC are the half of all the four angles; that is BAD and ADC are equal to two right angles: and therefore AB, CD B

C 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 prop. 34, book 1.


There are two cases of this proposition, one of which is not in the Greek text, but it 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, book 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 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 fifth and 7th propositions : for it is not worth while to relate his trifles 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.


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.


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,


case bis demonstration must be changed into another. now und finis


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 D this restriction, there might be three different cases of the proposition, as Campanus and others make,

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

G it is very easy to perceive, that DG being equal E 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.


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The proposition which is usually called the 5th postulate, or 11th axiom, by some the twelfth, 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 selfevident; but it may be demonstrated thus:


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.


A straight line cannot first come nearer to another straight line, and then go further from it, before it cuts A

B. it; and, in like manner, a straight line

D cannot go further from another straight F

H line, and then come nearer to it; nor can

G a straight line keep the same 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.

For example the straight line ABC cannot first come nearer to the straight line DE, as from the point A

B to the point B, and then, from the point A

С B to the point C, go further from the D same DE: and, in like manner the F


H straight line FGH 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.)


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 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

F nearer to AB from the point C to F: but because DB is greater than FE, the C

D straight line CFD 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 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.


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 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: therefore the angle ACD is G

F D equal (8. 1.) to the angle BDC: from any point E in AB draw EF unto CD, at right

G angles to AB: therefore by prop. 1. EF is equal to AC, or BD; wherefore, as has been just now shown, the angle ACF is equal to A

E B 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 angles 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.


If two straight lines which contain an angle be produced, there may be found in either side 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 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

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