The Elements of Euclid; viz. the first six books, together with the eleventh and twelfth. Also the book of Euclid's Data. By R. Simson. To which is added, A treatise on the construction of the trigonometrical canon [by J. Christison] and A concise account of logarithms [by A. Robertson].

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 be

ginning to learn the Elements, as not DM FG

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; 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; 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 Simpson 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 shews, that 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 be greater than it, as in the figure here annexed; in which case his demonstration must be changed into another.

PROPOSITION XIV.

To this is added, “ of the two sides DE, D 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.

Mr. Thomas Simpson, 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 shewn, that the point F falls below the line EG. This probably 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.

PROPOSITION XXIX.

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:

DEFINITIONS.

I. The distance of a point from a straight line, is the perpendicular drawn to it from the point.

II. 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 further from it, before it cuts it; and, in like manner, a straight line cannot go further from another straight line, and then 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.

D
F

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 point C, go further from the same DE: and, in like manner, 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.

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

PROPOSITION 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 BD.
Join AD, BC; and because, in the triangles CAB, DBA,

4. 1.

8. 1.

the sides CA, AB are equal to DB, BA, and the angle CAB equal to the angle DBA, the base BC is equal to the base AD and in the triangles ACD, BDC, the sides AC, CD are equal to BD, DC, and the base AD is equal to the base BC; therefore the angle ACD is equal to the angle BDC: from any point E in AB, draw EF to CD, at right angles to AB; therefore, (by Prop. 1.) EF is equal to AC, or BD; wherefore, as has been just shewn, 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 they are right angles *; 10 Def. 1. wherefore, also the angles ACD, BDC are right angles.

COROLLARY. 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.

* 15. 1.

4. 1.

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.

PROPOSITION 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 be equal to AE; and produce FE to H, and make EH equal to FE, and join HG. Because, in the triangles AEF, GEH, the two sides AE, EF are equal to the two GE, EH, each to each, and contain equal

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.
In the same manner
Prop. 2.) that is, to the double of FE.
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.

PROPOSITION 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.
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,

* 15. 1. * 4. 1.

that is EAB, is equal to the angle
FCH; wherefore, the angle AGF 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 together equal
to two right angles*: therefore, also AFG, AFH are equal to
two right angles, and consequently*, 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.

PROPOSITION 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 two right angles, AB and CD being produced, shall meet one another towards the points on which are the two angles, which are less than two right angles.

* 13. 1. 14. 1.

At the point C in the straight line CE, make * the angle 23. 1. 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 together equal to two right angles, the angles EAB, FCA are equal to two right angles. But, by the hypothesis, the angles EAB, ACD are together less than two

* 13. 1.

right angles; therefore 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 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

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The Elements of Euclid; viz. the first six books, together with the eleventh ...