VI. Def. 4. The fourth definition is strictly only applicable to a triangle, because no other figure has a point which can be exclusively called its vertex. The altitude of a parallelogram is the perpendicular drawn to the base from any point in the opposite side. VI. 2. The enunciation of this important proposition is open to objection, for the manner in which the sides may be cut is not sufficiently limited. Suppose, for example, that AD is double of DB, and CE double of EA; the sides are then cut proportionally, for each side is divided into two parts, one of which is double of the other; but DE is not parallel to BC. It should therefore be stated in the enunciation that the segments terminated at the vertex of the triangle are to be homologous terms in the ratios, that is, are to be the antecedents or the consequents of the ratios. It will be observed that there are three figures corresponding to three cases which may exist; for the straight line drawn parallel to one side may cut the other sides, or may cut the other sides when they are produced through the extremities of the base, or may cut the other sides when they are produced through the vertex. In all these cases the triangles which are shewn to be equal have their vertices at the extremities of the base of the given triangle, and have for their common base the straight line which is, either by hypothesis or by demonstration, parallel to the base of the triangle. The triangle with which these two triangles are compared has the same base as they have, and has its vertex coinciding with the vertex of the given triangle. VI. A. This proposition was supplied by Simson. VI. 4. We have preferred to adopt the term "triangles which are equiangular to one another," instead of " equiangular triangles," when the words are used in the sense they bear in this proposition. Euclid himself does not use the term equiangular triangle in the sense in which the modern editors use it in the Corollary to I. 5, so that he is not prevented from using the term in the sense it bears in the enunciation of VI. 4 and else where; but modern editors, having already employed the term in one sense ought to keep to that sense. In the demonstrations, where Euclid uses such language as "the triangle ABC is equiangular to the triangle DEF," the modern editors sometimes adopt it, and sometimes change it to "the triangles ABC and DEF are equiangular." Ia VI. 4 the manner in which the two triangles are to be placed is very imperfectly described; their bases are to be in the same straight line and contiguous, their vertices are to be on the same side of the base, and each of the two angles which have a common vertex is to be equal to the remote angle of the other triangle. By superposition we might deduce VI. 4 immediately from VI. 2. ;" VI. 5. The hypothesis in VI. 5 involves more than is directly asserted; the enunciation should be, "if the sides of two triangles, taken in order, about each of their angles... that is, some restriction equivalent to the words taken in order should be introduced. It is quite possible that there should be two triangles ABC, DEF, such that AB is to BC as DE is to EF, and BC to CA as DF is to ED, and therefore, by V. 23, AB to AC as DF is to EF; in this case the sides of the triangles about each of their angles are proportionals, but not in the same order, and the triangles are not necessarily equiangular to one another. For a numerical illustration we may suppose the sides of one triangle to be 3, 4 and 5 feet respectively, and those of another to be 12, 15 and 20 feet respectively. Walker. Each of the two propositions VI. 4 and VI. 5 is the converse of the other. They shew that if two triangles have either of the two properties involved in the definition of similar figures they will have the other also. This is a special property of triangles. In other figures either of the properties may exist alone. For example, any rectangle and a square have their angles equal, but not their sides proportional; while a square and any rhombus have their sides proportional, but not their angles equal. VI. 7. In VI. 7 the enunciation is imperfect; it should be, "if two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals, so that the sides subtending the equal angles are homologous; then if each....." The imperfection is of the same nature as that which is pointed out in the note on VI. 5. Walker. The proposition might be conveniently broken up and the essential part of it presented thus: if two triangles have two sides of the one proportional to two sides of the other, and the angles opposite to one pair of homologous sides equal, the angles which are opposite to the other pair of homologous sides shall either be equal, or be together equal to two right angles. For, the angles included by the proportional sides must be either equal or unequal. If they are equal, then since the triangles have two angles of the one equal to two angles of the other, each to each, they are equiangular to one another. We have therefore only to consider the case in which the angles included by the proportional sides are unequal. Let the triangles ABC, DEF have the angle at A equal to the angle at D, and AB to BC as DE is to EF, but the angle ABC not equal to the angle DEF: the angles ACB and DFE shall be together equal to two right angles. For, one of the angles ABC, DEF must be greater than the other; suppose ABC the greater; and make the angle ABG equal to the angle DEF. Then it may be shewn, as in VI. 7, that BG is equal to B BC, and the angle BGA equal to the angle EFD. Therefore the angles ACB and DFE are together equal to the angles BGC and AGB, that is, to two right angles. Then the results enunciated in VI. 7 will readily follow. For if the angles ACB and DFE are both greater than a right angle, or both less than a right angle, or if one of them be a right angle, they must be equal. VI. 8. In the demonstration of VI. 8, as given by Simson, it is inferred that two triangles which are similar to a third triangle are similar to each other; this is a particular case of VI. 21, which the student should consult, in order to see the validity of the inference. VI. 9. The word part is here used in the restricted sense of the first definition of the fifth Book. VI. 9 is a particular case of VI. 10. VI. 10. The most important case of this proposition is that in which a straight line is to be divided either internally or externally into two parts which shall be in a given ratio. The case in which the straight line is to be divided internally is given in the text; suppose, for example, that the given ratio is that of AE to EC; then AB is divided at G in the given ratio. Suppose, however, that AB is to be divided cxternally in a given ratio; that is, suppose that AB is to be produced so that the whole straight line made up of AB and the part produced may be to the part produced in a given ratio. Let the given ratio be that of AC to CE. Join EB; through C draw a straight line parallel to EB; then this straight line will meet AB, produce through B, at the required point. VI. II. VI. 14. This is a particular case of VI. 12. The following is a full exhibition of the steps which lead to the result that FB and BG are in one straight line. The angle DBF is equal to the angle GBE; add to each the angle FBE; [Hypothesis. therefore the angles DBF, FBE are together equal to the angles GBE, FBE. [Axiom 2. But the angles DBF, FBE are together equal to two right angles; [I. 13. therefore the angles GBE, FBE are together equal to two right angles; therefore FB and BG are in one straight line. [Axiom 1. [I. 14. VI. 15. This may be inferred from VI. 14, since a triangle is half of a parallelogram with the same base and altitude. It is not difficult to establish a third proposition conversely connected with the two involved in VI. 14, and a third proposition similarly conversely connected with the two involved in VI. 15. These propositions are the following. Equal parallelograms which have their sides reciprocally proportional, have their angles equal, each to each. Equal triangles which have the sides about a pair of angles reciprocally proportional, have those angles equal or together equal to two right angles. We will take the latter proposition. Let ABC, ADE be equal triangles; and let CA be to AD as AE is to AB: either the angle BAC shall be equal to the angle DAE, or the angles BAC and DAE shall be together equal to two right angles. [The student can construct the figure for himself.] Place the triangles so that CA and AD may be in one straight line; then if EA and AB are in one straight line the angle BAC is equal to the angle DAE. [I. 15. If EA and AB are not in one straight line, produce BA through and AF is equal to AE, therefore CA is to AD as AF is to AB. [Construction. [V. 9, V. 11. Therefore the triangle DA F is equal to the triangle BAC. [VI. 15. But the triangle DAE is equal to the triangle BAC. [Hypothesis. Therefore the triangle DAE is equal to the triangle DAF. [Ax. 1. Therefore EF is parallel to AD. [I. 39. Suppose now that the angle DAE is greater than the angle DAF. Then the angle CAE is equal to the angle AEF, [I. 29. [I. 5. and therefore the angle CAE is equal to the angle AFE, and therefore the angle CAE is equal to the angle BAC. [I. 29. Therefore the angles BAC and DAE are together equal to two right angles. Similarly the proposition may be demonstrated if the angle DAE is less than the angle DAF. VI. 16. This is a particular case of VI. 14. VI. 17. This is a particular case of VI. 16. There is a step in the second part of VI. 22 which requires examination. After it has been shewn that the figure SR is equal to the similar and similarly situated figure NH, it is added "therefore PR is equal to GH." In the Greek text reference is here made to a lemma which follows the proposition. The word lemma is occasionally used in mathematics to denote an auxiliary proposition. From the unusual circumstance of a reference to something following, Simson probably concluded that the lemma could not be Euclid's, and accordingly he takes no notice of it. The following is the substance of the lemma. If PR be not equal to GHI, one of them must be greater than the other; suppose PR greater than GH. Then, because SR and NH are similar figures, PR is to PS as GH is to GN. [VI. Definition 1. But PR is greater than GH, [Hypothesis. [V. 14. therefore PS is greater than GN. Therefore the triangle RPS is greater than the triangle HGN. [I. 4, Axiom 9. But, because SR and NH are similar figures, the triangle RPS is equal to the triangle HGN; which is impossible. Therefore PR is equal to GIT. [VI. 20. VI. 23. In the figure of VI. 23 suppose BD and GE drawn. Then the triangle BCD is to the triangle GCE as the parallelogram AC is to the parallelogram CF. Hence the result may be extended to triangles, and we have the following theorem, |