AXIOMS.

1. Things which are equal to the same are equal to one another.

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 unequal.

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.

8. Things which mutually agree with one another, are equal to one another.

9. The whole is greater than its part.

PROPOSITION I.

PROBLEM.

Upon a given finite right line to describe an equilateral triangle

Let AB be the given finite right line; it is required upon AB to describe an equilateral triangle. From the centre A with the distance AB de

scribe the circle BCD:a and again

a Post. 3.

from the centre B, with the dis

tance BA, describe the circle ACE, D and from the point c in which the circles cut one another, draw the right lines CA, CR, to the points A, B.

b

b Post. 1.

Therefore because A is the centre of the circle DBC, AC will be equal to AB. Again, because в is the centre Def. 15. of the circle CAE, BC will be equal to BA: but it has been shown that CA is equal to AB: therefore CA, CB, are each of them equal to AB. And things which are equal to the same are equal to one another. Whence CA is equal to CB; wherefore the three, CA, AB, BC, are equal to one another; and, consequently, the triangle ABC is equilateral, and it is described upon the given finite right line AB. 2. E. F.

PROPOSITION II.

PROBLEM.

From a given point to draw a right line equal to a given right line.

A

a

K B

D

☐ 1. 1.

G

Eb Post. 2.

Let a be the given point, and BC the given right line : it is required to draw from the point A a right line equal to the given right line BC. Draw the right line ac from the point A to c, and upon it describe the equilateral triangle DAC, and produce the right lines DA, DC, to E and F, and with centre c, and distance BC, describe the circle BGH. C Again, with centre D, and distance DG, Post. 3. describe the circle GKL: therefore because the point c

is the centre of the circle BGH, BC will be equal to Def. 15. CG. Again, because D is the centre of the circles GKL,

• Ax. 3.

a 2.1.

DL will be equal to DG and DA DC parts of them are equal: therefore the remainder AL is equal to the remainder CG. But it has been shown that BC is equal to CG. Wherefore each of them, AL, BC, is equal to CG. And things which are equal to the same thing are equal to one another. Whence AL is equal to BC. Therefore from a given point, AL has been drawn, &c. g. E. F.*

PROPOSITION III.

PROBLEM.

Two unequal right lines being given, to cut off from the greater a part equal to the less.

Let AB and c be two unequal given right lines of which AB is the greater: it is required to cut off from the greater, AB, a right line equal to

D

E B

c, the less. Draw from the point a a F right line, AD, equal to c;a and from the centre, A, with the distance AD, Post. 3. describe the circle DEF. And because A is the centre of the circle DEF, AD will be equal to AE. But AD is also equal to c. Therefore each of them, AE, C, will be equal to AD. Wherefore AE is also equal to c. Therefore two unequal right lines being given, &c.†

Ax. 1.

PROPOSITION IV.
THEOREM.

Q. E. F.

If two triangles have two sides equal to two sides, each to each; and have also one angle equal to one angle, viz. that which is contained by the equal right lines: then shall the base of the one be equal to the base of the other; and

*This proposition may be divided into a variety of cases according to the different positions of the point A, although the construction and demonstration will, in every respect, be the same. Proclus remarks that some performed it by taking the line AL with a pair of compasses; but he by no means approved of the method, as those who thus reason, he says, beg in the very beginning.

+ Some persons perform this proposition by taking the less line in the compasses, and with one leg in either extremity of the greater, cutting off with the other leg the part required: this, though correct in its operation, is certainly not geometrical, and would come rather under the class of postulates, than a demonstrable proposition.

one triangle equal to the other triangle; also the remaining angles of the one shall be equal to the remaining angles of the other, each to each, which are opposite to the equal sides.

Let there be two triangles, ABC, DEF, which have the two sides AB, Ac, equal to the two sides DE, DF, each to each; namely, the side AB equal to the side DE, and the side Ac equal to DF; also the angle BAC equal to the angle EDF. Then is the base BC equal to the base EF, and the triangle ABC equal to the triangle DEF; also the remaining angles equal to the remaining angles, R each to each, to which the equal

ДА

CE

sides are opposite; namely, the angle ABC to the angle DEF; and the angle ACB to the angle DFE.

For if the triangle ABC be applied to the triangle DEF, and the point A be put upon the point D, and the right line AB upon the right line DE, then shall the point в coincide with the point E, because AB is equal to DE. But AB coinciding with DE; the right line Ac shall also coincide with the right line Dr, since the angle BAC is equal to the angle EDF. Wherefore c will also coincide with F: for the right line ac is equal to the right line DF; but the point в coincides with the point E. Therefore the base BC will also coincide with the base EF. Because if the point в coinciding with the point E, and c with F; the base BC does not coincide with the base EF; two right lines would inclose a space which is impossible. Whence the base BC Ax, 10. coincides with the base EF, and also equal to it. Therefore the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it; also the remaining angles will coincide with the remaining angles, and equal to them,b viz., the angle ABC to the angle DEF, Ax. 8. and the angle ACB to the angle DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, &c. Q. E. D.

a

4.1.

PROPOSITION V.

THEOREM.*

The angles which are at the base of isosceles triangles are equal to one another; and the equal right lines being produced, the angles under the base shall be equal to one another.

F

BA

C

G

E

There

Let ABC be an isosceles triangle, having the side ab equal to the side Ac, and produce the right lines ab, AC, directly forward to D, E. Then is the angle ABC equal to the angle ACB, and the angle CBD to the angle BCE. For take in the line ED any point F: and from the greater AE cut off AG equal to AF the less: also join FC, GB. Therefore, because AF is equal to AG; and AB to AC; the two D FA, AC, are equal to the two GA, AB, each to each; and contain the common angle FAG. fore the base Fc is equal to the base GB, and the triangle AFC equal to the triangle AGB; also the remaining angles shall be equal to the remaining angles, each to each, viz., the angle ACF equal to the angle ABG; also the angle AFC to the angle AGB. And because the whole Ar is equal to the whole AG ; of which the parts AB, AC, are equal; the remaining part BF will also be equal to the remaining part CG. But it has been proved that Fc is equal to GB. Therefore the two BF, FC, are equal to the two CG, GB, each to each; and the angle BFC equal to the angle CGB: also their base BC is common. Whence the triangle BFC is equal to the triangle CGB; and the remaining angles equal to the remaining angles, each to each, to which the equal sides are opposite. Therefore the angle FBC is equal to the angle GCB; and the angle BCF to the angle CBG. Wherefore, because the whole angle ABG has been proved to be equal to the

FC

This theorem was discovered by Thales, for he is first said to have perceived and proved, that the angles at the base of every isosceles triangle are equal, and, after the manner of the ancients, to have called them similar. The latter part of it is not at all necessary in demonstrating the former; and it is affirmed by some geometers, amongst whom is Scarborough, that it is not Euclid's, but added by some one else; however this may be, the angles opposite the equal sides may be demonstrated without proving the equality of the angles under the base, as is evident by the very elegant and concise demonstration of Pappus, and indeed by many others.