But this is a known problem which may be solved from the given data.

Calling the intermediate problem a, and using the same notation as before, the analysis takes the form Pa H; and these steps reversed give the synthesis Ha P.

In general, however, the process is longer, and we require several successive substitutions before we arrive at a known problem which the data enable us to solve. The determination of these intermediate and auxiliary problems often requires considerable ingenuity, and only careful study and extensive practice of examples will give confidence and ensure success.

The solution of the following problem affords a fair example of the application of this method.

Given two concentric circles to draw a chord of the outer circle, which shall be trisected by the inner.

Particular Enunciation.-Let CDH and ABK be the concentric circles, having the common centre 0. It is required to draw a chord of the circle BAK, which shall be trisected by its intersection with the inner circle CDH.

Analysis.

Let ACDB be the required chord, which is trisected

in C and D.

Join OA, OC, F OD, OB. Then OCD, OAB are isosceles triangles

and the perpendicular OL bisects the bases CD and AB. Hence AL

B

H

K

=LB and CL=LD. Therefore taking equals from equals AC is always equal to DB, and if the chord is

=

drawn so that AC = CD, then DB will also be equal to CD. The problem P is there

fore resolved into (c) viz.:

To draw from a point A in the circumference of the outer circle a secant AD to the inner circle SO

F

B

that it may be bisected at the point of intersection C (c.)

Produce OC to F, making CF=OC. Then AD and OF will both be bisected in C, and the figure DFAO will be a // with FO and AD for its diagonals. Now OF=200=twice the radius of the smaller circle. Also OA= the radius of the larger circle, and OD=radius of the inner circle. The problem is therefore

reduced to this, viz.
Having given two sides and one diagonal
of a // to construct it. (b.)

Lastly, since AO=FD the three sides of the triangle ODF are known, and when this triangle is made the // can be completed and the problem solved.

But this is a known problem, viz.: prop. 22. Euc. Book I. (a).

We have therefore the following construction and proof.

Construction and Proof. Draw OD any radius of the inner circle. From centre D with radius equal to the radius of the outer circle, describe an arc, and with centre O and

radius=20D, cut this arc in F. Join FO

and FD.

Draw the radius OA parallel to FD and join FA. Then because OA is equal and parallel to FD, FA is equal and parallel to OD. [I. 33.] .. FAOD is a //m.

Lastly, draw the diagonal AD and produce to B.

Then the diagonals FO and AD bisect each other. [I. 26.] .. AC=CD and OC=CF. But OF=twice the radius of the inner circle, and hence the point C is in the inner circumference.

.*. AD is bisected by its intersection with the circle.

Join OB and drop the perpendicular OL. Then OAB, OCD are isosceles triangles, and their bases are bisected by OL.

:. AL=LB and CL = LD; and taking equals from equals AC=DB. But AC=CD. .. CD=AC=BD.

Hence, AB is trisected in C and D.

This example will show that while the method of Analysis and Synthesis is of great value, it still leaves much to the ingenuity of the student, and is neither an infallible method in all cases nor equally applicable to all. Further, it should be remarked that the final deduction from the analysis may be perfectly true, and yet we may not in all cases be able to retrace our steps and present a solution of the problem. We can only regard the method, therefore, as one which promises a fair chance of success in the majority of cases, and one which we ought to be prepared to try whenever the required solution is not immediately obvious.

There is, however, another general method known by the name of the "intersection of loci," which is of great value in the solution of problems, and which we now proceed to explain.

5. INTERSECTION OF LOCI.-A locus' in Geometry is the assemblage of all the points which fulfil a given condition. Thus, all the points in the circumference of a circle are at the same distance from the centre; and any point within the circle is at a less distance from the centre than any point in the circumference, while any point without the circle is at a greater distance from the centre. Hence the circumference of a circle is the assemblage of all the points whose distance from the centre is equal to the radius. Or in other words

I. The locus of all points which have a given distance from a fixed point, is the circumference of the circle described from the fixed point as a centre with the given distance as a radius.

Again, since parallel lines are equidistant from one another at all points, it can easily be shewn that,

II. The locus of all points which have a given distance from a fixed line, is the line drawn parallel to the fixed line, and at a distance from it equal to the given distance.

Again, if a perpendicular be drawn through the middle of the line joining two fixed points, it may be shewn by Euc. I. 4, that any point in this line is equally distant from the two fixed points, and indirectly by Euc. I. 8, that any point out of that line is unequally distant from the fixed points. Hence,

III. The locus of all points equally distant from two fixed points, is the perpendicular which bisects the line joining the points.

Lastly, if the angle made by two intersecting lines be bisected, it may be shewn by Euc. I. 26, that the perpendiculars from any point in this bisector upon the lines containing the angle are equal to one another; and hence,

IV. The locus of all points equally distant from two given lines, is the line which bisects the angle contained by them.

These are the most important geometrical loci which

depend upon the truths established in Euclid's first book. All other loci are of the same kind as these, i.e. they are either straight lines or circles. Hence, in the case of two loci, if they are both straight lines, their intersection will give but one point; but if one of them is a circle, their intersection will in general produce two points. Now in problems where it is required to determine a point which shall fulfil certain conditions, it frequently happens that the point belongs to two loci, one of them fulfilling one set of the conditions, and the other another set. In all such cases, if we find the locus (straight line or circle) which fulfils the one set of conditions only, and then the locus which fulfils only the other set, the required point will belong to both these loci, and will therefore be completely determined by their intersection.

Thus let it be required,

To find a point which shall be equally distant from three given points not in the same straight line.

1. Particular Enunciation.-Let A, B, C, be the three given points not in the same straight line.

2. Then, since the re

quired point is to be equally distant from A and B, it must lie in the locus of all points equally distant from two fixed points. But from iii. this is the perpendicular which bisects the distance between the points.

.. Join AB, and through the middle point P draw

P

S

PQ at right to AB. Then the required point lies in PQ.

3. Similarly if we bisect BC by the perpendicular SR, the required point must lie in SR. iii.

4. Hence, the point lies in both the lines SR and