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 A D /E B 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 C 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. † 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. 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 angle 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 BC did not coincide with EF, two straight lines would enclose a space, which (L. 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, 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 one another; their bases are likewise equal, and the triangles are equal to 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 (I. 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 (I. 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 BD 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, GB. F B A C G E Because (const.) AF is equal to AG, and (hyp.) AC to AB, the two sides FA, AC are equal to the two GÀ, 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 D 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 AF‡ divide the angle BAC * In this proposition, the hypothesis is, that the triangle under consideration haswo 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 into two equal parts. Then, in the triangles BAF, CAF, the side BA is equal (hyp.) to CA, and AF is common; therefore the two sides BA, AF, are equal to the two CA, AF, each to each, and the contained angles BAF, CAF are equal: wherefore (I. 4, part. 3) the remaining angles are equal, each to each, to which the equal sides are opposite; therefore, the angle ABF is equal to the angle ACF, and (I. 4, schol.) the exterior angles DBF, ECF, are D also equal. B C P E Cor. Hence every equilateral triangle is also equiangular. This is shown by taking, first, one side as base, and then another. PROP. VI. THEOR.*- If two angles of a triangle be equal to one another, the sides which subtend, or are opposite to, those angles, are also equal to one another. Let ABC be a triangle having the angle ABC equal to the angle ACB; the side AB is also equal to the side AC. D A For, if AB be not equal to AC, one of them is greater than the other let AB be the greater, and from it cut off (I. 3) BD equal to AC, the less, and join DC. Then, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each; and (hyp.) the angle DBC is equal to the angle ACB; therefore (I. 4, part 2) the triangle DBC is equal to the triangle ACB, the less to the greater; which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. Cor. B Hence every equiangular triangle is also equilateral. axiom, which ought not to be done without a good reason. In other respects this proof is perfectly satisfactory; and on account of its ease and simplicity, it may perhaps be preferred by those who are reading the Elements for the first time. This proposition may also be proved by conceiving the triangle BAC to be inverted, and to be applied to the space which it before occupied, so that the point A may retain its position, while AC will fall on AB; then will the point C fall on B, AB on AC, and B on C; and the angles ACB, ECB will coincide with ABC, DBC respectively, and (I. ax. 8) be equal to them. It may be remarked that Euclid's proof, when stripped of artifice, and referred to the principle of superposition, on which it, and every demonstration which has the fourth proposition for its basis, ultimately depends, is reduced to an inversion and mutual application such as that indicated above. This will appear evident from proving the equality of the triangles ABG, ACF by superposition. * This proposition is the converse of the fifth; that is, in the language of ic, the subject of this is the predicate of the fifth, and the predicate of this the surject of the fifth. Thus, it is proved in the fifth, that if two sides of a triangle be equal, the angles opposite to them are also equal; and in the sixth, that if two angles of a triangle be equal, the sides opposite to them are equal. This demonstration is indirect. To prove AB equal to AC, we suppose them, if possible, to be unequal. Then one of them, suppose AB, being considered the greater, we cut off a part DB, terminated at one of the equal angles, which we assume as equal, if possible, to AC; and joining CD, we prove, by means of the fourth proposition, that if the supposition were true, the triangle DBC would be PROP. VII. THEOR.*-Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. If it be possible, let there be two triangles ABC, ABD, upon the same base AB, and upon the same side of it, which have their sides CA, DA, terminated in the extremity A of the base equal to one another, and likewise their sides CB, DB, that are terminated in B. D Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal (I. 5) to the angle ADC. But the angle ACD is greater (I. ax. 9) than BCD; therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than BCD. Again, because CB is equal to DB, the angle BDC is equal (I. 5) to BCD; but it has been demonstrated to be greater than it; therefore the supposition, that AC is equal to AD, and likewise BC to BD, is false. A B E But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; therefore, because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other side of the base CD are equal (I. 5) to one another; but the angle ECD is greater (I. ax. 9) than BCD; wherefore the angle FDC is likewise greater than BCD: much more then is BDC greater than BCD. Again, because CB is equal to DB, the angle BDC is equal (I. 5) to BCD; but BDC has been proved to be greater than the same BCD; and therefore AC cannot be equal to AD, and at the same time BC to BD. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore on the same base, and on the same side of it, &c. A B equal in magnitude to ACB: but this being contrary to the 9th axiom, we conclude that AB and AC cannot be unequal, since the supposition that they are so, leads to an absurd conclusion. The student may exercise himself usefully in proving this proposition by producing CA through the vertex, and by supposing the whole line thus produced to be equal, if possible, to AB: and he may in like manner have two variations of the proof by supposing AC greater than AB. It would perhaps remove, in some degree, the difficulty which beginners often feel with respect to negative demonstrations, were they expressed hypothetically. Thus, in the present instance, we might proceed briefly in the following manner:For, if AB were not equal to AC, one of them, as AB, would be the greater. Then some part of AB, as BD, would be equal to AC; and, CD being joined, there would be two triangles DBC, ACB, in which there would be two sides and the contained angle of the one respectively equal to two sides and the contained angle of the other, and therefore (I. 4) the triangle DBC would be equal to ACB, which (I. ax. 9) is impossible. Therefore AB is not unequal to AC, that is, it is equal to it. It is often stated that this proposition affords the first instance in which the indirect method of proof is employed. This, however, is incorrect, as it is on this principle that the bases and the triangles themselves are proved to coincide in the fourth proposition, by showing that their failing to coincide would be contrary to the corollaries to the 3d and 5th definitions. This proposition is merely a lemma to the 8th, and may be omitted, if the second method of demonstrating that proposition be adopted. |