PLH greater than the angle PH L. This is obviously the case since the angle C, LH is equal to the angle C, H L (Euc. I., 5).

Or we might have noticed that the angle LCE is equal to the vertical angle PCA (Euc. I., 15) and therefore (hyp.) to QCE. Also, CL is equal to CQ and the triangle LCQ (here we draw in QE L) is isosceles, C E being the bisector of the angle contained by the equal sides. Hence C E is at right angles to QL and bisects QL in E. It is a very obvious consideration, at this point, that if we join C, L we shall have QC, L an isosceles triangle, C, Q being clearly equal to C, L (Euc. I., 4). Hence PC, and C, L together are equal to PC, and C, Q together. But PC,, C, L together are greater than P L (Euc. I., 20). Hence PC, and C, Q are together greater than PC, CQ together. We find, then, that our surmise is correct, for what we have proved for PC1, C1 Q1 can be proved equally well wherever C, may be taken. Thus the problem is solved. It is not necessary to give the synthetical statement of our solution, since this has already been given in the scholium to Example 8.

1

It may be argued that such tentative processes as we began with here are not mathematics. To this it is to be answered-first, that the art of guessing well is an important aid to the mathematician; and secondly, that we deal with our guesses by means of mathematical reasoning, and thus gain all the benefit available from mathematical processes.

But further, there are no laws for applying simple geometry that is geometry resembling Euclid's-to deductions; and therefore in many cases we have no choice but to make use of tentative methods.

VII. NON-EUCLIDIAN DEVICES.

We may remark in passing that there is no absolute necessity for restricting ourselves in all respects to Euclid's manner. Take as an instance his treatment of the famous pons asinorum. In dealing with this, as with all other propositions, he confines himself entirely to constructions which he has shown to be possible. Therefore, the following proof of the first part of the proposition would not be in his manner, though it would be difficult to find any flaw in the reasoning.

There must be some line which divides BAC (Fig. 17) into two equal angles.' Let A E represent this line. Then in the triangles BA E, CAE, BA is equal to AC (hyp.); A E is common; and the angle BA E is equal to the angle CA E. Therefore (by I., 4) the angle A B E is equal to the angle ACE.

Again, the following proof of both parts of the proposition is complete, though not in Euclid's

manner :

Conceive that the figure formed by the lines FK, FL, and G H (Fig. 18) is one that would coincide exactly with the figure formed by the lines A D, A E,

The assumption here is precisely the same in character as that made in defining a right angle.

and BC; FK coinciding with AD (Fig. 18), FL with A E, and GH with BC. Now conceive the figure FKL to be turned face downwards, and so applied to the figure A D E that FK may coincide with AE; then since the angle GFH is equal to the angle CA B, FL coincides with AD. Also since A B, AC are equal to each other, and also to FG, FH, the points G and H coincide with the points C and B, and G H with C B. Thus the angle

A AA

FIG. 17.

FIG. 18.

A B C coincides with and is equal to the angle F HG. But by our supposition, the angle ACB is equal to the angle FH G. Therefore the angle ABC is equal to the angle ACB. In like manner DBC coincides with GHL; but, by our supposition, BCE is equal to G H L. Therefore DBC is equal to BCE.

Or, we may produce A B and AC in Fig. 17, and

Here we assume as axiomatic the property which Simpson has attempted to prove in the corollary he has added to I., 12. He forgot, apparently, that Euclid had already (in Prop. 4 and elsewhere) assumed the property as self-evident, and that Prop. 12 itself cannot be solved on any other assumption.

conceive the part of the figure to the right of A E rotated round A E till it falls on the part to the left, and then show the perfect coincidence of the two portions.

In attacking geometrical deductions we are often compelled to assume in this way the existence of figures which are clearly conceivable, though we may not know precisely how to construct them, or though it may even be impossible to construct them by any of the ordinary geometrical processes. The following example of a problem in geometrical maxima and minima affords an instance:

Ex. 11.—A C B (Fig. 19) is part of a circle whose centre is at 0. The points P and Q lie without the circle. Determine under what conditions the sum of the distances PC and QC will be a minimum.

P

E

FIG. 19.

Here, guided by Examples 8, 9, to which the above is supposed to be given as a rider, we are readily led to the inference that PC and QC should be equally inclined to the tangent at C. Now there is no simple method of determining C so that this

relation may hold. But it is clear that there must be some position of C for which it holds. Conceive, then, that PC and QC are equally inclined to DCE, and let us inquire whether their sum is a minimum. Take any point F in A C, and join PF and QF. Then we have to show that PF and QF are together greater than PC, CQ. Let PF meet D C in G and join GQ. Then PG and GQ are together greater than PC and CQ (Example 9); and PF, FQ are clearly greater than P G, G Q (Euc. I., 20). Hence, à fortiori, PF, FQ are together greater than PC, CQ. Therefore the sum of PC and C Q is a minimum.

COR. Join CO; then the angle PCO is equal to the angle QCO, and we may express the relation deduced above thus:-

The sum of the lines drawn from any point without a circle to a point on the circumference will be a minimum when the two lines are equally inclined to the radius drawn to the last-named point.

The subject of geometrical maxima and minima is a wide one, but we shall content ourselves here by adding three in which areas are dealt with.

Ex. 12.-Two sides of a triangle being given, it is required to construct the triangle so that its area shall be a maximum.

Let A B, BC (Fig. 20) be the lengths of the given sides.

With centre B and radius BC describe the circle CDFE. Then if we draw any radius BD or BE (these radii accidentally omitted from the figure should