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.

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 externally 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 l through B, at the required point.

VI. II.

This is a particular case of VI. 12.

VI. 14. 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.

[I. 13.

But the angles DBF, FBE are together equal to two right angles; 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 A to F, so that AF may be equal to AE; join DF and EF. Then because CA is to AD as AE is to AB, [Hypothesis. and AF is equal to AE,

therefore CA is to AD as AF is to AB.

[Construction.

[V. 9, V.

II.

Therefore the triangle DAF 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,

and therefore the angle CAE is equal to the angle AFE,

[I. 29.

[I. 5.

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.
VI. 22.

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 GH, 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,

therefore PS is greater than GN.

[Hypothesis.

[V. 14.

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 GH.

[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,

triangles which have one angle of the one equal to one angle of the other, have to one another the ratio which is compounded of the ratios of their sides.

Then VI. 19 is an immediate consequence of this theorem. For let ABC and DEF be similar triangles, so that AB is to BC as DE is to EF; and therefore, alternately, AB is to DE as BC is to EF. Then, by the theorem, the triangle ABC has to the triangle DEF the ratio which is compounded of the ratios of AB to DE and of BC to EF, that is, the ratio which is compounded of the ratios of BC to EF and of BC to EF. And, from the 'definitions of duplicate ratio and of compound ratio, it follows that the ratio compounded of the ratios of BC to EF and of BU to EF is the duplicate ratio of BC to EF.

VI. 25. It will be easy for the student to exhibit in detail the process of shewing that BC and CF are in one straight line, and also LE and EM; the process is exactly the same as that in I. 45, by which it is shewn that KH and HM are in one straight line, and also FG and GL.

It seems that VI. 25 is out of place, since it separates propositions so closely connected as VI. 24 and VI. 26. We may enunciate VI. 25 in familiar language thus: to make a figure which shall have the form of one figure and the size of another.

VI. 26. This proposition is the converse of VI. 24; it might be extended to the case of two similar and similarly situated parallelograms which have a pair of angles vertically opposite.

We have omitted in the sixth Book Propositions 27, 28, 29, and the first solution which Euclid gives of Proposition 30, as they appear now to be never required, and have been condemned as useless by various modern commentators; see Austin, Walker, and Lardner. Some idea of the nature of these propositions may be obtained from the following statement of the problem proposed by Euclid in VI. 29. AB is a given straight line; it has to be produced through B to a point 0, and a parallelogram described on AO subject to the following conditions; the parallelogram is to be equal to a given rectilineal figure, and the parallelogram on the base BO which can be cut off by a straight line through B is to be similar to a given parallelo

gram.

VI. 32. This proposition seems of no use. Moreover the enunciation is imperfect. For suppose ED to be produced

through D to a point F, such that DF is equal to DE; and join CF. Then the triangle CDF will satisfy all the conditions in Euclid's enunciation, as well as the triangle CDE; but CF and CB are not in one straight line. It should be stated that the bases must lie on corresponding sides of both the parallels; the bases CF and BC do not lie on corresponding sides of the parallels AB and DC, and so the triangle CDF would not fulfil all the conditions, and would therefore be excluded.

VI. 33. In VI. 33 Euclid implicitly gives up the restriction, which he seems to have adopted hitherto, that no angle is to be considered greater than two right angles. For in the demonstration the angle BGL may be any multiple whatever of the angle BGC, and so may be greater than any number of right angles.

VI. B, C, D. These propositions were introduced by Simson. The important proposition VI. D occurs in the Meyáλn Σύνταξις of Ptolemy.

THE ELEVENTH BOOK.

IN addition to the first six Books of the Elements it is usual to read part of the eleventh Book. For an account of the contents of the other Books of the Elements the student is referred to the article Eucleides in Dr Smith's Dictionary of Greek and Roman Biography, and to the article Irrational Quantities in the English Cyclopædia. We may state briefly that Books VII, VIII, IX treat on Arithmetic, Book X on Irrational Quantities, and Books XI, XII on Solid Geometry.

XI. Def. 10. This definition is omitted by Simson, and justly, because, as he shews, it is not true that solid figures contained by the same number of similar and equal plane figures are equal to one another. For, conceive two pyramids, which have their bases similar and equal, but have different altitudes. Suppose one of these bases applied exactly on the other; then if the vertices be put on opposite sides of the base a certain solid is formed, and if the vertices be put on the same side of the base another solid is formed. The two solids thus formed are con tained by the same number of similar and equal plane figures, but they are not equal.

It will be observed that in this example one of the solids has a re-entrunt solid angle, see page 264. It is however true that