as that prescribed by arithmetic for finding the common divisor of two numbers; it needs, therefore, no other demonstration.

It is possible that, how far soever we continue the operation, we may never find a remainder which is contained an exact number of times in the preceding. In that case the two lines have no common measure, and are incommensurable. An instance of this will be seen

hereafter in the ratio of the diagonal to the side of the square.

In

such cases, then, the exact ratio in numbers cannot be found. The Ratio of Incommensurables is the limiting term of a series of ratios expressed in numbers. (See Introduction.).

PROBLEM XIX.

Two angles, A and B, being given, to find their common measure, they have one, and, by means of it, their ratio in numbers.

if

Describe, with equal radii, the arcs CD, EF, to serve as measures for the angles; proceed then in the comparison of the arcs CD, EF, as in the preceding problem; for an arc may be cut off from an arc of the same radius, as a straight line from a straight line.

C

B

D

E

F

We shall thus arrive at the common measure of the arcs CD, EF, if they have one, and thence to their ratio in numbers. This ratio will be the same as that of the given angles (Prop. XVII.); and if DO is the common measure of the arcs, DAO will be that of the angles.

SCHOLIUM. It may happen, also, that the arcs compared have no common measure. In that case, the remark with regard to Incommensurables in the Scholium of the previous proposition, applies.

GEOMETRICAL ANALYSIS.

The words Analysis and Synthesis are used in Geometry in a special sense. Synthesis is a mode of reasoning which begins with some established truth or something given, and ends with some new result, with something required either to be done or to be proved. Synthesis leads from principles to consequences.

Analysis, or the method of resolution, is the reverse of Synthesis,

or a method of reasoning from consequences to principles. The course of consecutive deduction is the same in both.

The synthetic method is pursued by Legendre as by Euclid in his elements of Geometry. They commence with certain assumed principles and definitions, and proceed to the solution of problems and the demonstration of theorems by successive inferences from them. The student has only to follow the reasoning by which the successive truths are established, without regard to the method of the discovery of these truths.

In Geometrical Analysis, we begin with assuming the truth of some theorem, or the solution of some problem; that is, assuming that what is required to be done has been effected; and we deduce from this assumption consequences which we can compare with known results, and thus test the truth of our assumption.

As Leslie has expressed it: "Analysis presents the medium of invention, while Synthesis naturally directs the course of instruction."

It is impossible to indicate any general and certain method for the demonstration of new theorems, or for the solution of problems by Analysis. Yet certain steps may be given which will render the investigation of new propositions easy and natural. These do not constitute so much a direct method of solution as a convenient way of searching for a suggestion. We give these steps separately for theorems and problems, remarking that the Geometrical Analysis is more extensively useful in discovering the solution of problems than for investigating the truth of theorems.

ANALYSIS OF THEOREMS.

Assume that the Theorem is true.

Construct the figure and examine any consequences that result from this assumption as a truth temporarily admitted, by the aid of other known truths respecting the figure. If any one of these consequences is known to be false, we have arrived at a reductio ad absurdum, which proves that the theorem is false. If a consequence can be deduced which coincides with some result already established, we start from this consequence, and endeavor, by retracing our steps, to give a synthetical demonstration of the theorem. This retracing our steps synthetically is essential, because a proposition may be false and yet furnish consequences that are true.

These directions are necessarily vague, however, as no certain rule can be given by which we can combine our assumption with truths already established. Nothing but experience, ingenuity, and a ready recollection of these truths will here avail the student. example to illustrate the steps indicated, we give the following

As an

THEOREM.

If two opposite sides of a parallelogram be bisected, and two lines be drawn from the points of bisection to the opposite angles, these two lines trisect the diagonal.

Let ABCD be a parallelogram of which the diagonal is AC. Let AB be bisected in E, and DC in F; also let DE, FB, be joined, cutting the diagonal in G, H. Then is

AC trisected in G and H.

Assume the theorem to be true; that is, AG = GH = HC, and draw EK parallel to GH.

=

D

E

F

K

II

=

B

Now (Book I., Prop. XXXIII.), ED and FB are parallel, and therefore EK GH, being parallels intercepted between parallels. And, hence, if the theorem is true, EK AG. Therefore, the two triangles AGE and EKB are equal, as a consequence of an assumption, as they have AE= EB, angle EAG=BEK by reason of the parallels, and AG = EK by assumption.

Now, let us see whether these triangles are equal, from the known relations of the parts of the figure.

We have AE=EB by construction, EAG=BEK, and AEG=EBK, by reason of the parallels; hence, these two triangles are equal (Book I., Prop. VIII.), and so the consequence deduced from our assumption agrees with previously established results.

=

We now retrace our steps and give the Synthesis. Draw EK parallel to AC. The two triangles AGE and EKB are equal, having AE EB, and the angles adjacent to these sides equal. Therefore, AG = EK. But EK GH, being opposite sides of a parallelogram. Hence, AG = GH. Similarly, by drawing through F a line parallel to GH, and meeting DE, we can prove GH = HC. Therefore, AG = GH = HC, and the diagonal AC is trisected in G and II.

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EXERCISES ON BOOK II.

THEOREMS.

1. A line from a point, A, without a circle, to the centre, O, meets the circumference in the points B and C. Show that AB is the shortest line, and AC the longest line, which can be drawn from A to the circumference. Prove, also, the same for a point, A, within

the circle.

2. Show that the shortest distance between two circumferences is measured on the line which joins the centres: taking, first, the circles exterior to each other; and, second, one interior to the other..

3. The chord through a point, A, in a circle, which is perpendicular to the radius through that point, is shorter than any other chord which can be drawn through A.

4. If two equal chords cut each other, the parts of the one are equal to the parts of the other respectively.

5. Conversely, if two chords which intersect each other have one part equal in each, the two chords are also equal.

6. If two tangents to a circle intersect each other, the parts of these tangents, from the point of intersection to the points of contact, are equal; and, also, the bisectrix of the angle of the two tangents passes through the centre of the circle.

7. The sum of two of the opposite sides of a circumscribed quadrilateral is equal to the sum of the two other opposite sides.

8. Prove the converse of Theorem 7. That is, if the sum of two opposite sides of a quadrilateral is equal to the sum of the two other opposite sides, then a circle tangent to three sides will be tangent to the fourth.

9. The square and rhombus are the only parallelograms in which a circle can be inscribed.

10. If three circles are tangent to each other externally, the tangents drawn through the three points of contact meet in one point.

11. If a circle be inscribed in a triangle, the distance from the vertex of any angle to the points of contact of its sides is equal to the semi-perimeter, minus the side lying opposite to this angle.

12. Suppose a circle, O, tangent to the two sides, AB and AC, of

an angle, BAC, at the points B and C; then draw a tangent, DE, to this circle, terminating in the two sides of the angle. Show that the perimeter of the triangle ADE is constant, whatever point of the arc BC we take as the point of contact of the tangent DE.

13. Show that if in the above figure we join D and E with the centre, O, the angle DOE is constant.

14. If through one of the points of intersection of two circumferences a line be drawn parallel to the line of centres, the sum of the two chords intercepted on this parallel is double the distance of the

centres.

The preceding theorems can be demonstrated without using the theorems of Book II. which relate to the measure of angles.

15. If two chords intersect on the circumference, the angle contained by one of them and the prolongation of the other (called an exscribed angle), is measured by half the sum of the arcs of the chords.

16. If two triangles have their angles equal, and are inscribed in the same circle, they are equal.

17. Three circumferences which pass respectively through the three vertices of a triangle and cut each other on the sides, all meet in the same point.

18. The circle which passes through the vertex of a triangle and through the adjacent feet of two perpendiculars, from the vertices on the opposite sides, passes also through the point of intersection of the perpendiculars.

19. The angles of the triangle formed by joining the three feet of the altitudes of a triangle, are bisected by these altitudes. (Use one of the circles described in the previous theorem, and the circle described on one side as a diameter, and compare inscribed angles.)

20. If two chords, AB, CD, intersect each other in a circle, the sum of the arcs, AC + BD, which they intercept on the circumference is equal to the sum of the arcs intercepted by two diameters parallel to these chords.

21. If the three vertices of an equilateral triangle, ABC, be joined to any point, P, of the circumscribed circle, by lines PA, PB, PC, then the line of junction, PA, which crosses the triangle, is equal to the sum of the other two. (Auxiliary Construction: On PA take PD PB, and join BD, in order to obtain triangles for comparison.)

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