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. 7. Things which are halves of the same, are equal. 8. Magnitudes which coincide with one another-that is, which exactly fill the same space-are equal to one another. 1 10 9. The whole is greater than its part. 10. All right angles are equal to one another. * 11. Two straight lines cannot be drawn through the same point, parallel to the same straight line, without coinciding with one another. 12. It is possible for another figure to exist, equal in every respect to any given figure. 13. If two things be equal, and a third be greater than one of them, it is also greater than the other.. 14. If two things be equal, and one of them be greater than a third, the other is also greater than the third. 15. If there be three magnitudes, of which the first is greater than the second, and the second is greater than the third, still more is the first greater than the third. DEFINITIONS OF TERMS. 1. A proposition is a portion of science, and is a theorem, a problem, or a lemma. 2. A theorem is a truth which is established by a demonstration. 3. A problem either proposes something to be effected, as the construction of a figure; or it is a question that requires a solution. 4. A lemma is a subordinate truth previously established, to be employed in the demonstration of a theorem, or the solution of a problem. 5. A hypothesis is an assumption made without proof, either in the enunciation of a proposition, or in the course of a demonstra. tion. 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. 1:12. 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, ABBREVIATIONS, &c. The signs + and are respectively 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 7 and < are respectively called greater and less. Thus, A 7 B, means that A is greater than B; and A ZB, that A is less than B. The sign =, called equal, is the sign of equality; thus, A = B 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, 3A means three times A; MA 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 AB? The terms Proposition, Problem, Corollary, Construction, Demonstration, Preparation, &c., are respectively abbreviated, Prop. Prob., Cor., Const., Dem., Prep., &c. 102 PROPOSITION 1. PROBLEM. To describe an equilateral triangle upon a given finite straight line.ca** Given, the straight line AB; it is required to describe an equilateral triangle upon it. (Const.) From the centre A, at the distance AB, describe (Postulate 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 and CB, to the points A and B; ABC shall be the equilateral triangle required. (Dem.) Because the point A is the centre of the circle BCD, AC is equal (Definition 17) 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 and CB are each of them equal to AB; therefore (Axiom 1) CA is equal to CB; wherefore CA, AB, and BC, are equal to one another; and the triangle ABC (Def. 26) 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. Given, the point A, and the straight line BC; it is required to draw from the point A a straight line equal to BC. (Const.) 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 and DB, to É 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 A L shall be equal to BC. (Dem.) Because the point B is the centre of the circle CGH, BC is equal (Def. 17) to BG; and because D is the centre of the circle GKL, DL is equal to DG, and DA, DB, parts of them, are equal, being sides of an equilateral triangle; therefore the remainder AL is equal to the remainder BG (Ax. 3). But it has been shewn 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. de Given, AB and C two straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less. (Const.) From the point A draw (I. 2) the straight line AD equal to C; and from the centre A, and with the radius AD, describe (Post. 3) the circle DEF. (Dern.) Because A is the centre of the circle DEF, AE is equal to AD; but the straight line C is likewise equal to AD; whence AE and C are 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. PROPOSITION IV. THEOREM. 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. Given, ABC and DEF, two triangles which have the two sides AB and AC, equal to the two sides DE and DF, each to each, namely, AB to DE, and AC to DF; and the angle BAC equal to the angle EDF, to prove that 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 /G ABC to the angle DEF, and the angle ACB to DFE. (Const.) 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, (Dem.) the point B shall coincide with the point Ē, because AB is equal to DE; and AB coinciding with DÉ, 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 (Def. 3), and shall bė. H 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. CoR.-If the equal sides AB and DE be produced to G and H, the angles GBC and HEF below the base will also be equal. For BG and EH will coincide, and therefore the angle GBC is equal to the angle HEF. PROPOSITION V. THEOREM. 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. Given, ABC an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB and AC be produced to D and E, to prove that the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. Jor (Const.) In BD take any point F, and from AE, the greater, cut off AG equal (I. 3) to AF, the less, and join FC and GB. (Dem.) Because AF is equal to AG, and AB to AC, the two sides FA and AC are equal to the two GA and AB, each to each; -- and they contain the angle FAG common to the two triangles AFC and AGB; therefore the base FC is equal (I. 4) to the base GB, and the triangle AFC to the triangle AGB; and the 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 and FC are equal to the two CG and GB, each to each ; but the angle BFC is equal to the angle CGB; wherefore the triangles BFC and 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 (I. 4, cor.) the angle ABC is equal to the angle ACB. po COR.-Hence every equilateral triangle is also equiangular.. Scholium.—This proposition may be very simply demonstrated by bisecting the vertical angle Å by a line cutting the base. Then (I. 4) there are two triangles equal in every respect, and therefore the angles at the base are equal. And by the corollary OUT |