modern geometers: It seems not to be properly placed Boox I. among the axioms, as indeed it is not self-evident; but it may be demonstrated thus:

DEFINITION Ì.

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 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 fur

C

ther from it, before it cuts it;

B

and, in like manner, a straight D

E

line cannot go further from an

F

H

other straight line, and then

G

come nearer to it; nor can 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 to the point B, and then, from the point B to the

A.
D.

point C, go further from the same F

DE: And, in like manner, the

B

C See the

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 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 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

с

D

F

Boos I 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 C 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

E

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.

PROP. II.

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 BĎ.

Join AD, BC; and because, in the triangles CAB, DBA, CA, AB are equal to DB, BA, and the angle CAB equal 4. 1. 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 an

C

8. 1. gle ACD is equal to the angle
BDC: From any point E in AB
draw EF unto CD, at right angles
to AB; therefore by Prop. 1. EF A

F D

is equal to AC, or BD; wherefore, 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 10 Def. 1. EFC and EFD are equal, 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 right angle; but ACD is also a right angle; wherefore the angles

ACD, ACG, are equal to one another, which is impossible. Book I. Therefore BD is equal to AC; and by this proposition

BDC is a right angle.

PROP. 3.

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 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 equala angles, the angle GHE is therefore equal 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, KG is equal to FH, by Cor. Pr. D

2. that is, to

N

P

F

K

E

H

B

C

M

L

the double of
FE. In the same manner if AG be produced to L, so
that 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. 4.

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.

Book I.

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 FAG, a 15. 1. thata is EAB, is equal to the angle 4. 1. FCH; wherefore the angle AGF

E

is equal to CHF, and AFG to the
angle CFH: To these last add the
common angle AFH; therefore
the two angles AFG, AFH are
equal to the two angles CFH,

HFA, which two last are equal

13. 1. together to two right angles : therefore also AFG, AFH are equal

14. 1. to two right angles, and consequentlyd 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. 5.

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 two 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.

a 23. 1. At the point C, in the straight line CE, makea the angle ECF equal to the angle EAB, and draw to AB the straight line CG at right angles to CF: Then, because the angles ECF, EAB are equal to one another, and that the angles ECF, FCA are toge

13. 1. gether equal to two right angles, the an

E

gles EAB, FCA are

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. Book I. Let this point be H, and draw HK perpendicular to CF, meeting AB in L: And because AB, CF contain equal angles with AC on the same side of it, by Prop. 4. AB and CF are at right angles to the straight line MNO, which bisects AC in N, and is perpendicular to CF: Therefore by Cor. Prop. 2. CG and KL, which are at right angles to CF, are equal to one another: And HK is greater than CG, and therefore is greater than KL, and consequently the point H is in KL produced. Wherefore the straight line CDH, drawn betwixt the points C, H, which are on contrary sides of AL, must necessarily cut the straight line AB.

PROP. XXXV. B. I.

THE demonstration of this Proposition is changed, because if the method which is used in it was followed, there would be three cases to be separately demonstrated, as is done in the translation from the Arabic; for, in the Elements, no case of a Proposition that requires a different demonstration ought to be omitted. On this account, we have chosen the method which Mons. Clairault has given, the first of any, as far as I know, in his Elements, page 21, and which afterwards Mr. Simpson gives in his, page 32. But whereas Mr. Simpson makes use of Prop. 26. B. I. from which the equality of the two triangles does not immediately follow, because, to prove that, the 4th of B. I. must likewise be made use of, as may be seen in the very same case in the 34th Prop. B. I. it was thought better to make use only of the 4th of B. I.

PROP. XLV. B. I.

THE straight line KM is proved to be parallel to FL from the 33d Prop. whereas KH is parallel to FG by construction, and KHM, FGL have been demonstrated to be straight lines. A corollary is added from Commandine, as being often used.

PROP. XIII. B. II.

In this proposition only acute angled triangles are men- Book IL tioned, whereas it holds true of every triangle; and the demonstrations of the cases omitted are added: Commandine and Clavius have likewise given their demonstrations of these cases.