6. A corollary is a consequence easily deduced from one or more propositions.

7. A scholium is a remark on one or more propositions, which explains their application, connection, limitation, extension, or some other important circumstance in their nature. 8. A demonstration is a process of reasoning, and is either direct or indirect.

9. A direct demonstration is a regular process of reasoning from the premises to the conclusion.

10. An indirect demonstration establishes a proposition, by proving that any hypothesis contrary to it, is contradictory or absurd; and it is, therefore, sometimes called a reductio ad absurdum.

11. The enunciation is the statement or expression of the proposition; the particular enunciation is its statement in reference to a particular figure or figures.

12. The construction is an operation by which lines are drawn or points determined according to certain conditions. 13. The data or premises of a proposition are the relations or conditions granted or given, from which new relations are to be deduced, or a construction to be effected.

EXPLANATION OF SIGNS.

The signs and are called plus and minus. The former indicates addition; thus A + B is the sum of A and B, and A+B+C is the sum of A, B, and C: the latter indicates subtraction; thus, A B is the excess of A above. B; so A+B-C is the excess of A + B above C.

The signs and are called greater and less. Thus AB means that A is greater than B; and A A is less than B.

B, that

The sign =, called equal, is the sign of equality; thus AB implies that A is equal to B.

The small letters a, b, c, m, n, p, q, &c. are commonly used to denote numbers. A number placed before any quantity serves as a multiplier to it: thus, 3 A means three times A; m A means m times A, or A taken as often as there are units in m.

The square described on a line A is sometimes expressed by A called A square; or, if AB be the line, by AB2.

To describe an equilateral triangle upon a given finite straight line.

Let AB be the given straight line; it is required to describe an equilateral triangle upon it.

From the centre A, at the distance AB, describe (Postulaté 3) the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and

from the point C, in which the circles cut one another, draw the straight lines (Post. 1) CA, CB, to the points A, B ; ABC shall be an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is equal (Definition 14) to AB; and because the point B is the centre of the circle ACE, BC is equal to BA; but it has been proved that CA is equal to AB; therefore CA, CB, are each of them equal to AB; therefore (Axiom 1) CA is equal to CB; wherefore CA, AB, BC, are equal to one another; and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB.

PROPOSITION II. PROBLEM.

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

Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC. From the point A to B draw (Post. 1) the straight line AB; and upon it describe (I. 1) the equilateral triangle DAB, and produce (Post. 2) the straight lines DA, DB, to E and F; from the centre B, with the radius BC, describe (Post. 3) the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL shall be equal to BC. Because the point B is the centre of the

circle CGH, BC is equal (Def. 14) to BG; and because D is the centre of the circle GKL, DL is equal to DG, and

C

DA, DB, parts of them, are equal; therefore the remainder AL is equal to the remainder BG (Ax. 3): But it has been shown that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC.

PROPOSITION III. PROBLEM.

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

Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less.

From the point A draw (I. 2) the straight

but

line AD equal to C; and from the centre A, and with the radius AD, describe (Post. 3) the circle DEF; and because A is the centre of the circle DEF, AE shall be equal to AD; the straight line C is likewise equal to AD; whence AE and Care each of them equal to AD; wherefore the straight line AE is equal to C (Ax. 1), and from AB, the greater of two straight lines, a part AE has been cut off equal to C, the less.

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 one another, their bases, or third sides, shall be equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, namely, those to which the equal sides are opposite. Or, if two sides and the contained angle of one triangle be respectively equal to those of another, the triangles are equal in every respect.

Let ABC, DEF, be two triangles which have the two sides AB, AC, equal to the two sides DE, DF, each to each, namely, AB to DE, and AC to DF; and the angle BAC

equal to the angle EDF, the base BC shall be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles, to which the equal sides are opposite, shall be equal, each to each; namely, the angle

ABC to the angle DEF, and the angle ACB to DFE. For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB upon DE, the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F, because AC is equal to DF: but the point B coincides with the point E; wherefore the base BC shall coincide with the base EF (Cor. Def. 3), and shall be equal to it. Therefore, also, the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it; and the remaining angles of the one shall coincide with the remaining angles of the other, and be equal to them; namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other side of the base shall also be equal.

Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC, be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE,

In BD take any point F, and from AE, the greater, cut off AG equal (I. 3) to AF, the less, and join FC, GB. Because AF is equal to AG, and AB to AC, the two sides FA, AC, are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal (I. 4) to the base GB, and the tri- F angle AFC to the triangle AGB; and the n/

B

remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal sides are opposite; namely, the angle ACF to the angle ABG, and the angle AFC to the angle AGB: And because the whole AF is equal to the whole AG, and the part AB to the part AC, the remainder BF shall be equal (Ax. 3) to the remainder CG; and FC was proved to be equal to GB; therefore the two sides BF, FC, are equal to the two CG, GB, each to each; but the angle BFC is equal to the angle CGB; wherefore the triangles BFC, CGB, are equal, and their remaining angles are equal, 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. Now, since it has been demonstrated that the whole angle ABG is equal to the whole ACF, and the part CBG to the part BCF, the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC: And it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base.

COR.-Hence every equilateral triangle is also equiangular.

Scholium.-This proposition may be very simply demonstrated by bisecting the vertical angle A by a line cutting the base. Then (1.4) there are two triangles equal in every respect, and therefore the angles at the base are equal. After Proposition 13 is proved, it could then be easily shown that the angles on the other side of the base are equal. It is evident that some line will bisect the vertical angle; and although the method of doing it is not known till Problem 9 be solved, yet this does not affect the truth of the proposition.

This proposition may also be easily demonstrated by supposing another isosceles triangle, having its two sides and vertical angle equal to those of ABC (Ax. 12). For if ABC, DEF, in Proposition 4, be these triangles, it is proved, as in Proposition 4, that angle B = E; but as in this case the two sides CA, AB, are also respectively equal to ED, DF, it follows that angle C = E (I. 4); and as B and C are each E, therefore B = C.