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2. If equals be added to equals, the wholes are equal.
3. If equals be taken from equals, the remainders are equal.
4. If equals be added to unequals, the wholes are unequal.
5. If equals be taken from unequals, the remainders are umequal.
6. Things which are double of the same are equal to one another.
7. Things which are halves of the same, are equal to one another.

II. The Five Axioms, which apply especially to magnitude, are peculiarly AXIOMS OF GEOMETRY:

8. Magnitudes which coincide with one another,--that is, which exactly fill the same space,_are equal.

“ This Axiom is properly the definition of Geometrical equality. If all the parts and boundaries of one figure fall upon and cover all the parts and

boundaries of another figure, the two figures are equal. The name of Super-
position is given to this process, which when actually performed, is the proof to

the senses ; when only conceived of, demonstration to the mind.
9. The whole is greater than its part.
10. Two straight lines cannot enclose a space.
11. All right angles are equal to one another.
This may be illustrated by sup-

À
posing that equal circles
have been drawn from Band
E, the vertices of the right
angles : the fourth part of
each circle, or the arcs CA

CIG
FD, will be the measure of
fG

F each right angle ; but

B

E
the circles are equal; there-
fore the fourth parts of them
must be equal ; and conse-
quently the angles measured
by those equals must them-

H selves be equal. 12. If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these two straight lines, being continually produced, shall at length meet upon that side on which the angles are less than two right angles. “ Two st. lines which intersect one another, cannot be both parallel to the same st.

line."- PLAYFAIR.
An illustration, will make the
axiom plainer. The
с

A
angles DGH and GHF are
less than two right angles,

D
and CD, EF meet in K. II.
The angles CGH, and GHE,

I
K к

II
being less than two right
angles, the lines CD and EF
H)

E meet in L; i.e., on the side

E

F of AB on which the

angles E
are less than two right
angles.

B
Lines converge, when they approach nearer and at last meet in a point K: but lines

diverge, when, setting out from a point L, they recede more and more.

I.

C

L

3.

PROP. 1.-PROB.
To describe an equilateral triangle upon a given finite straight line.

The SOLUTION requires Pst. 3, Pst. 1.–The DEMONSTRATION.-Def. 15, Ax.1.
E.1 Dat. Given the st. line AB;

2 Quæs. on it to describe an equil. A. C.1 by Pst. 3. From centre A, with AB,

C describe O BCD; 2

from centre B, with BA,

describe O ACE; 3 1. and from C, or F, where D

the Os cut, draw CA, CB,

or FA, FB; 4.Sol.

then A ABC, or AFBA,

shall be an equil A. D.1 by C. 1. Def.15.:: A is the centre of O BCD, AC = AB; 2 C.2. Def. 15. · B

of O ACE, BC = BA; 3 D.1,2. Ax. 1. But AC, BC,"each AB, . . AC= BC. 4 D. 1, 2, 3. Thus AB, BC, and AC, sides of a A, are equal to one

another. 5 Rec. Therefore the A ABC is equilateral and drawn on AB. Q. E. F.

A APPLICATION.-By an equil. triangle, BDE, or CFG, made of wood or metal, to ascertain an inaccessible distance, BA or CA. At B place the triangle so that A can be seen along BD; without change of position, look along BE, and set out a line in continuation of BE; carry the triangle along BC, until by looking along the edge DE, the same object A will appear in a st. line with CF: then A, B, and C will be the angular points D of an equil. triangle; and as the sides are equal, by measuring from B to C, the distance from B to A, or from C to A, will be ascertained.

Ac

PROP. 2.-PROB.
From a given point to draw a st. line equal to a given st, line.

Sol.-Pst. 1, P. 1, Pst. 2, Pst. 3.-DEM.-Def. 15, Ax. 3, Ax, 1.
E.1 Dat. Given the · A and the st. line BC;
Quæs. from A to draw a st.line equal to BC.

K
C.1 byPst. 1. From B the extremity of
BC, draw a line to the · A;

H 2 P. 1. on AB construct an equil.

A BDA;
Pst. 2. lengthen DB and DA in-
definitely to E and F;

B
Pst. 3. from centre B, with BC,

describe the O CHG; 5

and from centre D, with

DG, describe the O GLK; 6 Sol. then the st. line AL, from

F
A, is equal to the given st.
line BC.

3

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D.1byC. 4, 5. B is the centre of the O CGH, and D of the O GLK; 2

Def. 15. BG BC, and DL DG. 3 C, 2. But the part DA = the part DB; 4 Sub. and taking these equals from the equals DL, and DG; 5 Ax. 3. .:. the remainder AL = the remainder BG; 6 D. 2.Ax. 1.) but BG = BC, .. AL = BC. 7 Rec. Wherefore from the given · A there

has been drawn a st. line AL = the

given st. line BC. Q. E. F. SCHOLIUM.-When the given point is out of the given line, or its production, this problem admits of eight cases;

н pat if the given point is in the given line, or its production, of only four cases.

of the eight cases we will give the diagram for one, in which the given point A is joined to C, the other extremity of the given line BC.

When the proposition admits of it, the learners improvement is much promoted by varying the mode of the solution. He may solve any of the twelve cases for himself.

Ei

A

PROP. 3.-PROB.

From the greater of two given st. lines to cut off' a part equal to the less

SOL.-P. 2, Pst. 3.-DEM.-Def. 15, Ax. 1.

E. I Dat. Given the st. lines AB and C, AB being the greater of the two;

2 Quæs. it is required to cut off from AB a part equal to C the less.

A

C. I by P. 2 From the draw a st. line

AD= C; 2 Pst. 3. and from A, with radius AD,

describe the O DEF; 3 Sol. then the st. line AE, cut off

from AB, the less. D. 1 by C. 2. ... A is the centre of the O

E

B

FED,

2 D. 15. .. the st. line AE = the st.

line AD:
31 C. l. But the st. line AD = the

given st. line C,
Ax. l. consequently, the st. line AE,

cut off from AB, = the st. line C. 5. Rec. Therefore from AB, the greater, has been cut off, &c. Q.E.F.

· APP. 1.- A less line AD may be made equal to a greater AB, by describing a circle GBH, with the radius AB, and producing AD to meet the circle in H; then AH will equal AB.

2.-To construct a Scale of Equal Parts; Take AB of indefinite length towards B, and let C be the given st. line or part, that is to be cut off from AB : from AB cut off a part equal to C, as AE; then from EB another part equal to C, as EF; and so on; the parts in

AB are each equal to C and to one another; and AB is a scale of equal parts; for the radii AD, ED, FD, GD, being each equal to C, therefore AE, EF, FG, and GB are all equal.

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3.-On the same principle we take a lina KL, and from K set off ten equal parts, as in KO; then from O set parts each equal to KO: if the parts between K and are tenths, the parts 1, 2, 3, 4, 5 will be units; but if the parts between K and O are units, then the parts numbered 1, 2, 3, 4, 5 will be tens, namely, 10, 20, 30, 40, 50. By a scale of this kind, the comparative lengths of lines may be readily measured.

4.–To use a Scale of Equal Parts intelligently, we should understand the nature o Representative Values. A miniature is representative of the human face; and a map may be representative of an immense tract of the heavens. The lines in the miniature and in the map are in due proportion to those in the face and in the expanse of heaven; and thus they possess a representative value,-they are not the actual distances, but they stand for them. The inch on a scale may indicate a mile, or a thousand miles of distance; but if each portion of a mile, if each mile, or thousand miles, is given of the same relative size, then the map is a true representation ; stretch out all its parts in due proportion, and at last, by superposition, it would actually cover every point of the wide surface of which it is but the mark or outline, See Green's Euclid, practically applied, notes to Prop. 3, 4, 20, 26, 31, 32, book I, and Prop. 5, 6, 18, book VI.

5. By such a use of the Scale of Equal Parts, and of the Scale for Angular Magnitude, 've can construct figures that are a true index of the positions and real distances of cities, mountains and seas, and in some respects of the constellations of heaven.

PROP. 4.-THEOR.—(Important.)

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to each other ; they shall likewise have their bases, or third sides, equal; and the two triangles shall be equal, and their angles shall be equal, each to each, viz., those to which the equal sides are opposite.

А A

DEM.-Ax. 10, Ax. 8.
E. 1 Hyp. 1. In the two A8 ABC, DEF, let AB

DE, and AC= DF;
2
2. Also let the included

< BAC = the included / EDF; 3 Conc. 1. then the base BC = the base

AA

EF;

4 5

2. and A ABC = A DEF;
3. also, _ ABC= _ DEF,

and Z ACB = Z DFE.

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8 Con. sup.

D. 1 by Superp. Applying A ABC to A DEF, so that · A is on · D,

and AB on DE; 2 D.1, H.1

.: A B coincides with and is equal to D E, 3 Ax. 8.

the · B shall coincide with the · E; 4 D.2,3,H.2 Again, ':. AB coincides with DE, and < BAC= _ EDF, 5 Conc.

the line AC shall fall on the line DF: 6 H.1,&Conc. But AC being DF,.. the · C shall fall on the . F; 7 D. 3, 6. and · B falling on · E, and C on F, the line B C falls on

the line EF: Forif, though B falls on E,and Con F,BCdoes not fall on EF,

then two st. lines will enclose a space; 9. Ax. 10. But this is impossible ; 10 Ax. 8. .. the base BC does coincide with and = the base EF: 11 D.2. 5, 6, 10 And A B falling on and being equal to D E; A C to D F;

and BC to EF, 12 Ax. 8. ..A ABC coincides with and A DEF. 13 Hyp. 1, D.10 Also, since D E coincides with A B, and EF with BC, 14 Ax.8, Cone. the L A B C shall coincide with and equal / DEF. 15 Hyp.1, D. 7. And in a similar way / ACB=LD FE. :6 Rec. Wherefore, if two triangles have two sides, &c. Q.E.D.

Sch.-1, This being the first Theorem, it is exclusively proved by means of the Axioms. The converse of the 8th Axiom is assumed ; namely, that if the magnitudes are equal, not merely if they are equivalent, they will coincide.

2. The equality spoken of is equality of the sides and of the angles. Triangles may be equal in area, though the sides and angles of the one are not equal to the sides and angles of the other. When the sides and angles mutually coincide, the triangles are named equal triangles ; when their areas only are equal, equivalent triangles.

App.—This proposition contains the first of the criteria by which to infer the equality of triangles, and is applied to various uses : as

1st. In all parts of Geometry to establish the equality of triangles.
2nd. To ascertain an inaccessible distance, as A B, the breadth of a lake.

Take the angle at C formed by the lines from the extremities, A and B, of the inaccessible distance; and measure un the distances CA and CB. The representative Values of these measure- A

JBE

F ments must now be taken from a Scale of Equal Parts, and drawn on paper, or on any plane surface : thus, draw a st. line DF of an indefinite length, and at D, by aid of the graduated semicircle, form an angle FDE, equal to the angle BCA : from a scale of equal parts represent the distance from A to C, in proper proportion, by the line DE, and the distance from B to C, by DF ; consequently, on a principle established in the Sixth Book, Prop. 4, and which we now assume as a Lemma,-“that the sides about similar triangles are proportional,--the line BF will represent the distance from A to B : and if to the same scale we apply the line EF, that distance on the scale wlll be the representative measurement of the actual distance AB,

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