с D PROPOSITIONS. PROPOSITION I. PROBLEM.* _To describe an equilateral triangle on 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 (I. postulate 31) the circle BCD, and from the centre B, at the distance BA, describe (I. post. 3) the circle ACE; and from the point C, in which the circles cut one another, draw (I. post. 1) the straight lines CA, CB to the points A, B: ABC is the equilateral triangle required. Because the point A is the centre of the circle BCD, AC is equal (I. definition 30) to AB; and because the point B is the centre of the circle AČE, BC is equal (I. def. 30) to BA. But it has been proved that CA is equal to AB; therefore, CA, CB are each of them equal to AB: but things which are equal to the same are equal (I. axiom 1) to one another; therefore CA is equal to CB; wherefore CA, AB, BC are equal to one another; and the triangle ABC is therefore (I. def. 17) equilateral, and it is described upon the given straight line AB: which was required to be done. Scholium. If straight lines be drawn from A and B, to F, the other point in which the circles cut one another, it would be proved * The words in which a proposition is expressed, are called its enunciation. If the enunciation refer to a particular diagram, it is called a particular enunciation: -otherwise, it is a general one. · A demonstration is a series of arguments which establish the truth of a theorem, or of the solution of a problem. Demonstrations are either direct or indirect. The direct demonstration commences with what has been already admitted, or proved, to be true, and from this deduces a series of other truths, each depending on what precedes, till it finally arrives at the truth to be proved. In the indirect or negative demonstration, or as it is also called, the reductio ad absurdum, a supposition is made which is contrary to the conclusion to be established. On this assumption, a demonstration is founded, which leads to a result contrary to some known truth; thus proving the truth of the proposition, by showing that the supposition of its contrary leads to an absurd conclusion. The drawing of any lines, or the performing of any other operation that may be necessary in a proposition, is called the construction. In this proposition, the first paragraph is the general enunciation; the second, the particular one; the third, the construction; and the fourth, the demonstratior. + That is, to describe an equilateral triangle, which shall have given a straight line as one of its sides. The word finite is employed to show that the line is not of unlimited length, but is given in magnitude, as well as position. The student should be accustomed to point out the data in problems and the hypotheses in theorems. In this problem, a straight line is given, and it is required to describe on it an equilateral triangle. In the references, the Roman numerals denote the book, and the others, when no word is annexed to them, indicate the proposition;-otherwise the latter denote a definition, postulate, or axiom, as specified. Thus, III. 16 means the sixteenth proposition of the third book; and I. ax. 2, the second axiom of the first book. So also hyp. denotes hypothesis, and const. construction. § By a scholium is meant a note or observation. in the same manner that AFB is an equilateral triangle. Hence on any straight line two equilateral triangles may be described one on each side of it.* A B F E PROP. II. PROB.–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 A a straight line equal to BC. From the point A to B draw (I. post. 1) the straight line AB; and upon it describe (I. 1) the equilateral triangle DAB, and produce (I. post. 2) the straight lines DB, DA, to E and F. From the centre B, at the distance BC, describe (I. post. 3) the circle CEH, and from the centre D, at the distance DE, describe (1. post. 3) the circle EFG. AF is equal to BC.I Because the point B is the centre of the circle CEH, BC is equal (I. def. 30) to BE; and because D is the centre of the circle EFG, DF is equal (I. def. 30) to DE; and DA, DB, parts of them are equal: therefore the remainder AF is equal (I. ax. 3) to the remainder BE. But it has been shown, that BC is equal to BE; wherefore AF and BC are each of them equal to BE ; and things that are equal to the same are equal (I. ax. 1) to one another; therefore the straight line AF is equal to BC. Wherefore from the given point A a straight line AF has been drawn equal to the given straight line BC: which was to be done. PROP. III. PROB. From the greater of two given straight lines to cut off a part equal to the less.Ş * It will be shown in the 10th proposition of the 3d book, that two circles can cut each other in only two points; and hence there can be only two equilateral triangles on a straight line. In the practical construction it is sufficient to describe small arcs intersecting each other in C or F. + Here the data are a point and a straight line. # A straight line may be drawn from the point A, to either extremity of BC, and on either of the lines thus drawn, two equilateral triangles may be constructed, Hence there may be four straight lines drawn, any one of which will be such as is required in the problem. To this there is an exception, when A is at either extremity of BC, or in its continuation, as in either case it will readily appear that only two such lines, can be drawn by the construction here given. It is plain also that if A be in BC, the equilateral triangle is to be described on either of the parts into which BC is divided at that point. In practice, every person will solve this problem by opening the compasses to the distance between B and C; and then, one point being placed at A, the other will mark out the point F on any line drawn from A. This method, however, though greatly preferable in practice, could not in strictness be adopted by Euclid, as it is not derived from the postulates or the first proposition. It is the same in fact as assuming this proposition as an additional postulate. In a science, however, which is strictly demonstrative, the fewer first principles that are assumed the better; and though it may sometimes be convenient, and may save time, to dispense with strict rigour, it is always satisfactory to know that the object in view may be attained without any departure from geometrical accuracy. § Here the data are two straight lines. In practice, what is done in this propo A B Let AB and C be the two given straight lines, of which 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 line AD equal to C; and from the centre A, at the distance AD, describe (I. post. 3) the circle DEF: AE is the part required. Because A is the centre of the circle DEF, AE is equal (I. def. 30) to AD; but the straight line C is likewise equal (const.) to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal (I. ax. 1) to C, and from AB, the greater of two straight lines, a part AE has been cut off equal to the less : which was to be done. Schol. By drawing (I. 2) from either extremity of C, a straight line equal to AB, the line C might be produced, till it and the part added would be equal to AB. Also, by producing AB through A, to meet the circumference, a line would be obtained equal to C and AB together. PROP. IV. THEOREM.— If two triangles have two sides of the one equal to two sides of the other, each to each ;* and have also the angles contained by those sides equal to one another: (1) they have likewise their bases, or third sides, equal; (2) the two triangles are equal; # and (3) their other angles are equal, each to each, viz. those to which the equal sides are opposite. Let ABC, DEF be two triangles which have the two sides AB, sition and in the scholium, will be effected simply by means of the compasses, as was pointed out in the preceding proposition. * The meaning of the expression each to each, or respectively, which is used in the same sense, will be known from its application here. Were this expression wanting, the meaning might be merely that the sides AB, AC are together equal to DE, DF; while, when taken separately, they might be either equal or unequal. With the limiting expression, however, the meaning is, that AB is equal to DE, and AC to DF. In such cases, the lines or magnitudes must be taken in the same order. Thus, it would be improper in the present case to say, that AB, AC are equal to DF, DE, each to each. f This expression shows the meaning which Euclid attaches to the base of a triangle. It is the third side as distinguished from the other two, whether that is the side on which the triangle stands or not. # That is, they have equal areas or surfaces, the area of a superficial figure being the space which it contains. $ This enunciation might be more briefly expressed thus:- If two triangles have two sides, and the contained angle of the one respectively equal to two sides and the contained angle of the other; they have likewise their remaining sides equal, and their remaining angles equal, each to each, viz. those which are similarly situated; and their areas are equal. In this proposition, the hypothesis is, that there are two triangles which have two sides and the contained angle in one of them, equal respectively to two sides and the contained angle in the other; and it is proved that if this be so, the triangles must be in every respect equal. AA н AC equal to the two sides, DE, DF, each to each, viz. AB to DE, and AC to DF; and the angle BAC equal to the angle EDF: then (1) the base BC is equal to the base EF; (2) the triangle ABC to the triangle DEF; and (3) the other angles, to which the equal sides are opposite, are equal, each to each, viz. the angle ABC to the argle DEF, and the angle ACB to DFE. For, if the triangle ABC be applied to 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 (hyp.) AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because (hyp.) the angle BAC is equal to the angle EDF; wherefore also the point ( shall coincide with the point F, because (hyp.) AC is equal to DF. But the point B coincides with E; wherefore the base BC shall coincide with the base EF; because, the point B coinciding with E, and C with F, if BĆ did not coincide with EF, two straight lines would enclose a space, which (I. def. 3, cor.) is impossible. Therefore the base BC shall coincide with the base EF, and (I. ax. 8) be equal to it. Wherefore the whole triangle ABC shall coincide (I. def. 5, cor.) with the whole triangle DEF, and (I. ax. 8) be equal to it; and the other angles of the one shall coincide with the remaining angles of the other, and (I. ax. 8) be equal to them, viz. the angle ABC to DEF, and the angle ACB to DFE.* Therefore, it two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another; their bases are likewise equal, and the triangles are equal, and their other angles to which the equal sides are opposite, are equal, each to each : which was to be demonstrated. Schol. If the equal sides AB, DE be produced through B and E, the extremities of the bases, the exterior angles CBG, FEH The following observations may assist the beginner in understanding the proof of this important proposition. When AB is applied to DE, it will coincide with it by the third definition. Also B will coincide with E, because AB is equal to DE. If B fell between D and E, AB would be less than DE; but if B fell on the continuation of DE through E, AB would be greater than DE. Again, the angles A and D are equal; that is, the opening of the lines AB, AC is equal to that of DE and DF. Hence, AB coinciding with DE, AC must coincide with DF. For if AC fell beyond DF, the angle A would be greater than D; but if AC fell between DE and DF, A would be less than D. The point C is then shown to coincide with F, in the same manner in which B was shown to coincide with E. Now, B and C coinciding with E and F, the lines between them must coincide so as to exclude space; for if they did not in every part coincide, a space would be enclosed between them: which (I. def. 3, cor.) is impossible. Hence these lines (1. ax. 8) are equal. The triangles must also coincide entirely: for, since the three sides of the one triangle coincide respectively with those of the other, if the triangles did not coincide internally, a space would be enclosed between two plane surfaces, which (I. def. 5, cor.) is impossible. The triangles are therefore (1. ax. 8) equal. Lastly, since the straight lines BA, BC coincide at the same time with ED, EF, the openings between them must be equal ; that is, the angles B and E must be equal: and C and F must be equal for a like reason. When a proposition is demonstrated by supposing one figure to be applied to another, it is said to be proved by the method of superposition. are equal to one another; for the lines which contain them coincide, when the one triangle is applied to the other. In like manner, it would appear that the angles formed by producing AC and DF through C and F are equal. PROP. V. THEOR.—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 are also equal.* Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the sides AB, AC be produced to D and E; the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE. In BDt take any point F, and from AE the greater, cut off (I. 3) AG equal to AF the less, and join (I. post. 1) FC, ĜB. Because (const.) AF is equal to AG, and (hyp.) AC to AB, the two sides FA, AC are equal to the two GA, AB, each to each; and they contain the angle A common to the two triangles AFC, AGB: therefore (I. 4, part. 1) the base FC is equal to the base GB; and the remaining angles of the one are equal (I. 4, part. 3) to the remaining angles of the other, each to each, to which the equal sides are opposite; therefore the angle AFC is equal to the angle AGB. And because the whole AF is equal (const.) to the whole AG, of which the parts AB, AC are (hyp.) equal; the remainder BF is equal (I. ax. 3) to the remainder CG; and FC has been proved to be equal to GB: therefore, in the triangles FBC and GCB, the two sides BF, FC are equal to the two CG, GB, each to each: and the angle BFC has been proved to be equal to CGB; wherefore (I. 4, part. 3) their remaining angles are equal, each to each, to which the equal sides are opposite; therefore the angle FBC is equal to the angle GCB, and (I. 4, schol.) the angle ABC is equal to the angle ACB, which are the angles at the base of the triangle ABC: and it has been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore the angles at the base, &c.; which was to be demonstrated. OTHERWISE :-Let the straight line AFị divide the angle BAC B G D E * In this proposition, the hypothesis is, that the triangle under consideration has yo of its sides equal; and on this hypothesis it is to be proved that the angles opposite to those sides are equal. The demonstration of this proposition given by Euclid is generally felt by beginners to be one of the most difficult in the Elements. The one here given proceeds on the same general principle, but is shorter and easier. + The proof would be equally easy were F and G taken in AB and AC. Through the point A an infinite number of straight lines may be drawn, and it is plain that there is one of these which divides the angle BAC into two equal parts. In assuming the truth of this, however, we virtually employ an additional |