371. THEOREM [Po.4].-If a quadrilateral is inscribed in

a circle, then the intersection of its diagonals, and the two intersections of pairs of opposite sides, are the vertices of a triangle which is self-conjugate to the circle.

Required to prove that A XYZ (Fig. 508) is self-conjugate.

M

FIG. 508.

FIG. 509.

= 1

1. § 358.

Proof. Since AC, BD, XZ are the diagonals of a complete quad., ... AC is divided harmon. by BD and XZ, i. e. {AYCL} = Project this range from the point Z on to the lines AX, DX, thus {AMBX} = {DNCX}

Thus M and N both lie on the polar of X;

that is, YZ is the polar of X.

=

- 1.

Po. 3.

Similarly YX is the polar of Z.

Hence A XYZ is self-conjugate to the circle.

§ 368.

372. THEOREM [Po.5].-If a point lies outside a circle its

polar is the chord of contact-that is the line joining the points

of contact of the two tangents from the given point.

Required to prove that the polar of P is LM (Fig. 509).

Also LM OP. Hence LM is the polar of P.

373. THEOREM [Po.6].-If a variable chord passes through a fixed point, then the locus of the intersection of the tangents at the extremities of the chord is the polar of the point.

Required to prove (Fig. 510) that, if LM is a variable chord through a fixed point P, then the locus of Q is the polar of P.

semicircle on AB which is toward CD in M, and the semicircle on CD

which is toward AB in N. Join MN and produce it to meet the circles again in P and Q respectively. Then PA, PB, QC, QD will form a square.

=

Or thus:-Join BD, draw AK 1 BD and produce to F so that AF BD. Then CF is the direction of one side of the square.

2. Required to draw a triangle similar to a given triangle having two of its vertices on OP, OQ, and the other vertex at C1 (Fig. 511):

0

FIG. 511.

B,

Two circles can be drawn through A and B touching XY. The required point P is one of the two points in which these circles touch XY.

EXERCISES CXXXIII.*

1. Show that the angle between two given lines is equal to the angle subtended at the centre by the join of their poles with respect to any given circle.

2. Show that the pole of a diameter is the point at infinity in the direction perpendicular to the diameter.

3. If a triangle be self-conjugate with regard to a circle, the centre of the circle is the orthocentre of the triangle and the triangle must be oblique-angled.

4. Construct the polar of a given point using the ruler only.

5. Construct the pole of a given straight line using the ruler only. 6. Draw tangents to a circle from a given point using the ruler only. 7. If D, E, F be three points in the sides of a triangle ABC, and perpendiculars be drawn at these points to the sides in which they lie, the necessary and sufficient condition that these perpendiculars should be concurrent is BD2 - DC2 + CE2 – EA2 + AF2 — FB2 = 0. Deduce that the following sets of lines are concurrent :

(1) The perpendiculars to the sides at the middle points.

(2) The perpendiculars to the sides from the opposite vertices.

8. If the perpendiculars from D, E, F to the sides of the triangle ABC respectively meet in a point, so also do the perpendiculars from A, B, C to the sides of the triangle DEF. [Use Ques. 7.]

9. ABCD is a cyclic quadrilateral, and AB, DC produced meet in P ; PE, PF are drawn perpendicular to AD, BC respectively; prove that AE: ED CF : FB.

10. ABC is a triangle right-angled at A. A variable straight line FD perpendicular to the hypotenuse meets AB, AC respectively in D and F. Find the locus of the intersection of CD and BF.

11. About a given quadrilateral circumscribe a quadrilateral similar to another given quadrilateral.

12. P is a point inside a parallelogram ABCD. If PBA = L PDA, prove that PAD = = PCD. [Through P draw parallels to sides.] 13. APB is an arc of a circle and P is any point on the arc; prove that the perpendicular from P to AB is a mean proportional between the perpendiculars from P to the tangents at A and B.

14. O is a fixed point. P moves along a fixed straight line not passing through 0. PO is at right-angles to OP and proportional to it. Find the locus of Q.

15. Given three parallel lines, construct a triangle of given species whose vertices lie on these lines.

16. Construct a triangle of given species, one of whose vertices is at a fixed point while the two others lie one on each of two given circles. 17. Given three concentric circles construct an equilateral triangle having its vertices one on each of these circles.

CHAPTER XXVI.*

IRRATIONAL NUMBERS.

375. Incommensurables. We have already defined two quantities of the same kind as commensurable if it is possible to find a third quantity which is contained an exact number of times in each of them (see § 133), i. e. if it is possible to find a common measure of the two quantities; and we have also pointed out that if two quantities are commensurable the ratio of one to the other is either an integer or a vulgar fraction.

Any two quantities of the same kind, however, are not necessarily A commensurable. For example—

The diagonal and the side of a square are incommensurable quantities.

Proof. In sq. ABCD, let AE bisect DAC, and let EF be I AC.

Then it is easily proved that AD = AF, DE EF FC, and EC is the diagonal of a square whose side is DE.

=

If possible let some length L be a common measure of DC and AC, so that DC = pL, ACqL, where p and q are integers.

Then

and

But q

=

B

E

C

FIG. 512.

DE FC AC AD = qL - pL = (q − p)L,
EC DC DE = pL-(q- p)L = (2p − q)L.

p and 2pq are integers; hence L is also a common measure of DE and EC, i. e. of the side and diagonal of a square whose side is DE. Moreover, it is easy to show that DE < DC.

By repeating the above argument it could then be shown that L is also a common measure of the side and diagonal of a square whose side is less than DE, and therefore <(1)2DC; and so on.

Thus ultimately L must be a common measure of the side and diagonal of a square whose side is <()"DC, where n may be any integer. But by increasing n sufficiently (1)"DC may be made less than any assignable magnitude, and therefore less than L.

But L cannot be a measure of a quantity less than itself. Hence there is no common measure to the side and diagonal of a square.

376. The Concept of Number.-If two quantities are commensurable, we can express one as an exact fraction of the other for example, if a third quantity is contained exactly p times in the first and exactly q times in the second, then the first quantity is p/q of the second. If, however, the two quantities are incommensurable we cannot express one as an exact fraction of the other : that is to say, if one of these two quantities is the unit of measurement, then the other cannot be expressed as an exact number (integral or fractional) of these units. Thus the arithmetical concept B of number is theoretically incomplete.

A

FIG. 513.

Let us look more carefully into this question. Suppose the line AB (Fig. 513) contains 3 units of length, and suppose that a point P travels from A to B. Then we conceive the distance AP as increasing continuously from zero to 3 units: that is to say, the distance AP passes through all possible magnitudes between zero and AB; and yet if we use only the arithmetical concept of number we have no means of expressing the distance AP accurately in terms of the unit of length whenever P is in such a position that AP is incommensurable to the unit of length. Accordingly we require somehow to extend our conception of number in such a way that number (like magnitude) may be supposed to increase continuously from any one value to any other.

Thus we must assume that numbers exist which cannot be accurately expressed by integers or fractions, and that any two incommensurable quantities have a ratio expressed by means of such a number.

377. Irrational Numbers.—All numbers which can be expressed by integers or fractions (proper or improper) are called rational numbers, and all numbers which cannot be so expressed are called irrational.

The object of this chapter is to give a logical account of irrational numbers. It will be necessary

(i) to explain how an irrational number is specified; (ii) to explain how irrational numbers can be sub

The existence of such numbers can be proved from simpler premises -see Russell's Principles of Mathematics,—but the discussion is far too difficult for an elementary book.