6th Axiom, as any thing in geometry. But, if still the removal of the triangle from its place be considered as creating a difficulty, and as inelegant, because it involves an idea, that of motion, not essential to geometry, this defect may be entirely remedied, provided that, to Euclid's three postulates, we be allowed to add the following, viz. That if there be two equal straight lines, and if any figure whatsoever be constituted on the one, a figure every way equal to it may be constituted on the other. Thus if AB and DE be two equal straight lines, and ABC a triangle on the base AB, a triangle DEF every way equal to ABC may be supposed to be constituted on DE as a base. By this it is not meant to assert that the method of describing the triangle DEF is actually known, but merely that the triangle DEF may be conceiv ed to exist in all respects equal to the triangle ABC. Now, there is no truth whatsoever that is better entitled than this to be ranked among the Postulates or Axioms of geometry; for the straight lines AB and DE being every way equal, there can be nothing belonging to the one that may not also belong to the other.

On the strength of this postulate the fourth Proposition is thus demonstrated.

If ABC, DEF be two triangles, such that the two sides AB and AC of the one are equal to the two ED, DF of the other, and the angle BAC, contained by the sides AB, AC of the one, equal to the angle EDF, contained by the sides ED, DF of the other; the triangles ABC and EDF are every way equal.

AA

B

G

E

F

On AB let a triangle be constituted every way equal to the triangle DEF; then if this triangle coincide with the triangle ABC, it is evident that the proposition is true, for it is equal to DEF by hypothesis, and to ABC, because it coincides with it; wherefore ABC, DEF are equal to one another. But if it does not coincide with ABC, let it have the position ABG; and first suppose G not to fall on AC; then the angle BAG is not equal to the angle BAC. But the angle BAG is equal to the angle EDF, therefore EDF and ABC are not equal, and they are also equal by hypothesis, which is impossible. Therefore the point G must fall upon AC; now, if it fall upon AC but not at C, then AG is not equal to AC; but AG is equal to DF, therefore DF and AC are not equal, and they are also equal by supposition, which is impossible. Therefore G must coincide with C, and the triangle AGB with the triangle ACB. But AGB is every way equal to DEF, therefore, ACB and DEF are also every way equal. Q. E. D,

By help of the same postulate, the 5th may also be very easily

demonstrated.

Let ABC be an isosceles triangle, in which AB, AC are the equal sides; the angles ABC, ACB opposite to these sides are also equal. Draw the straight line EF equal to BC, and suppose that on EF the triangle DEF is constituted every way equal to the triangle ABC, that is, having DE equal to AB, DF to AC, the angle EDF to the angle BAC, the angle ACB to the angle DFE, &c.

Α

A A

B

C

Then, because DE is equal to AB, and AB is equal to AC, DE is equal to AC; and for the same reason, DF is equal to AB. And beeause DF is equal to AB, DE to AC, and the angle FDE to the angle BAC, the angle ABC is equal to the angle DFE, (4: 1.). But the angle ACB is also, by hypothesis, equal to the angle DFE; therefore the angles ABC, ACB are equal to one another. Q. E. D.

Thus also, the 8th proposition may be demonstrated independently of the 7th.

Let ABC, DEF be two triangles, of which the sides AB, AC are equal to the sides DE, DF each to each, and also the base BC to the base EF; the angle BAC is equal to the angle EDF.

On BC, which is equal to EF, and on the side of it opposite to the triangle ABC, let a triangle BGC be constituted every way equal to the triangle DEF, that is, having GB equal to DE, GC to DF, the angle BGC to the angle EDF, &e. : join AG.

Because GB and AB are each equal, by hypothesis, to DE, AB and GB are equal to one another and the triangle ABG is isosceles. Wherefore also (5. 1.) the angle BAG is equal to the angle BGA. In the same way, it is shown that AC is equal to GC, and the angle CAG to the angle CGA. Therefore adding equals to equals, the two angles BAG, CAG together are equal to the two angles BGA, CG 4 together, that is, the whole angle BAC to the whole BGC. But the angle BGC is, by hypothesis, equal to the angle EDF, therefore also the angle BAC is equal to the angle EDF. Q. E. D.

Such demonstrations, it must, however, be acknowledged trespass against a rule which Euclid has uniformly adhered to throughout the Elements, except where he was forced by necessity to depart from it. This rule is, that nothing is ever supposed to be done, the manner of doing which has not been already taught, so that the construction is derived either directly from the three postulates laid down in the beginning, or from problems already reduced to those postulates. Now, this rule is not essential to geometrical demonstration, where, for the purpose of discovering the properties of figures, we are certainly at liberty to suppose any figure to be constructed, or any line to be drawn, the existence of which does not involve an impossibility. The only use, therefore of Euclid's rule is to guard against the introduction of impossible hypothesis, or the taking for granted that a thing may exist which in fact implies a contradiction; from such suppositions, false conclusions might, no doubt, be deduced, and the rule is therefore useful in as much as it answers the purpose of excluding them. But the foregoing postulatum could never lead to suppose the actual existence of any thing that is impossible; for it only assumes the existence of a figure equal and similar to one already existing, but in a different part of space from it, or having one of its sides in an assigned position. As there is no impossibility in the existence of one of these figures, it is evident that there can be none in the existence of the other.

PROP. VII.

Dr. Simson has very properly changed the enunciation of this proposition, which, as it stands in the original, is considerably embarrass. ed and obscure. His enunciation, with very little variation, is retained here.

PROP. XXI.

It is essential to the truth of this proposition, that the straight lines drawn to the point within the triangle be drawn from the two extremities of the base; for, if they be drawn from other points of the base, their sum may exceed the sum of the sides of the triangle in any ratio that is less than that of two to one. This is demonstrated by Pappus

Alexandrinus in the 3d Book of his Mathematical Collections, but the demonstration is of a kind that does not belong to this place. If it be required simply to show, that in certain cases the sum of the two lines drawn to the point within the triangle may exceed the sum of the sides of the triangle, the demonstration is easy, and is given nearly as follows by Pappu, and also by Proclus, in the 4th Book of his Commentary on Euclid.

Let ABC be a triangle, having the angle at A a right angle: let D be any point in AB; join CD, then CD will be greater than AC, hecause in the triangle ACD the angle CAD is greater than the angle ADC. From DC cut off DE equal to AC; bisect CE in F, and join BF; BF and FD are greater than BC and CA.

Because CF is equal to FE, CF and FB are equal to EF and FB, but CF and FB are greater than BC, therefore EF and FB are greater than BC. EF and FB add ED, and to BC add AC, which is equal to ED by construction,

To

C

E

D

and BF and FD will be greater than BC and CA. Q. E. D. It is evident, that if the angle BAC be obtuse, the same reasoning may be applied.

This proposition is a sufficient vindication of Euclid for having demonstrated the 21st proposition, which some affect to consider as selfevident; for it proves, that the circumstance on which the truth of that proposition depends is not obvious, nor that which at first sight it is supposed to be, viz. that of the one triangle being included within the other. For this reason I cannot agree with M. Clairaut, that Euclid demonstrated this proposition only to avoid the cavils of the Sophists. But I must, at the same time, observe, that what the French Geometer has said on the subject has certainly been misunderstood, and, in one respect, unjustly censured by Dr. Simson. The exact translation of his words is as follows: "If Euclid has taken the trouble to demonstrate, that a triangle included within another has the sum of its sides less than the sum of the sides of the triangle in whịch it is included, we are not to be surprised. That geometer had to do with those obstinate Sophists, who made a point of refusing their assent to the most evident truths," &c. (Elemens de Geometrie par M. Clairaut. Pref.)

Dr. Simson supposes M. Clairaut to mean, by the proposition which he enunciates here, that when one triangle is included in another, the sum of the two sides of the included triangle is necessarily less than the sum of the two sides of the triangle in which it is included, whether they be on the same base or not. Now this is not only y not Euclid's proposition, as Dr. Simson remarks, but it is not true, and is directly contrary to what has just been demonstrated from Proclus. But the fact seems to be, that M. Clairaut's meaning is entirely different, and that he intends to speak not of two of the sides of a triangle, but of all the three; so that his proposition is, “that when one

triangle is included within another, the sum of all the three sides of the included triangle is less than the sum of all the three sides of the other," and this is without doubt true, though I think by no means self-evident. It must be acknowledged also, that it is not exactly Euclid's proposition, which, however, it comprehends under it, and is the general theorem, of which the other is only a particular case. Therefore, though M. Clairaut may be blamed for maintaining that to be an Axiom which requires demonstration, yet he is not to be ac cused of mistaking a false proposition for a true one.

PROP. XXII.

Thomas Simson in his Elements has objected to Euclid's demonstration of this proposition, because it contains no proof, that the two circles made use of in the construction of the Problem must cut one another; and Dr. Simson on the other hand, always unwilling to acknowledge the smallest blemish in the works of Euclid, contends, that the demonstration is perfect. The truth, however, certainly is, that the demonstration admits of some improvement; for the limitation that is made in the enunciation of any Problem ought always to be shown to be necessarily connected with the construction of it, and this is what Euclid has neglected to do in the present instance. The defect may easily be supplied, and Dr. Simson himself has done it in effect in his note on this proposition, though he denies it to be necessary.

Because that of the three straight lines DF, FG, GH, any two are greater than the third, by hypothesis, FD is less than FG and GH, that is, than FH, and therefore the circle described from the centre F, with the distance FI) must meet the line FE between F and H; and, for

the like reason, the circle described from the centre G at the distance GH, must meet DG between D and G, and therefore, the one of these circles cannot be wholly within the other. Neither can the one be wholly without the other, because DF and GH are greater than FG; the two circles must therefore intersect one another.