8. The circle is the limiting form of the ellipse when the distance between the foci is indefinitely diminished.

For, in this case, S and H coincide with C.

9. If a chord of a circle pass through a fixed point, the tangents at its extremities will intersect on a fixed straight line. In fig., p. 87, supposing the curve a circle, let T be the intersection of tangents at the extremities of that chord which is bisected in the fixed point 0.

Let o be the middle point of any other chord through O, and t the intersection of tangents at its extremities.

But

Then

CO.CT= (radius)" = Co. Ct.

Hence a circle goes round Oot T.

Therefore

[Euc. III., 36, Cor.

OTt+ Oot = two right angles. [Euc. III., 22.

LOot

= a right angle.

[Euc. III., 3. Therefore OTt is a right angle, and Tt a fixed straight line, since T is by construction a fixed point.

10. A chord of a circle, drawn from any point, is cut harmonically by the point, the curve, and the polar of the point.

Using the construction of Prop. XVII., p. 88, and remembering that chords of a circle are perpendicular to the diameters which bisect them, we have

=

tp.tp' (tangent from t)2
=to.tc,

by similar right-angled triangles.

[Euc. III., 36.

Also, since the angles at O and c are right angles, a circle

11. To construct an equilateral triangle corresponding to a given triangle.

In the definition on p. 143, the ordinates may be parallel to any fixed straight line and the ratio may be any given ratio. Let SPH be the given triangle.

[fig., p. 102. Describe the equilateral triangle SgH and let gP meet HS in G.

Let the ordinates be parallel to PG, and let PG: gG be the constant ratio.

Then the straight line gH corresponds to PH. [Prop. I.,p.143. Similarly, gS corresponds to PS.

Hence the equilateral triangle gHS corresponds to the given triangle.

12. To construct a square corresponding to a given parallelogram.

Let ABCD be the given parallelogram.

Describe the square ABC'D' and let D'D meet AB in N. Then, if the ordinates be parallel to D'D and the constant ratio be that of ND to ND', the square ABC'D' will correspond to the given parallelogram, for AD' corresponds to AD and BC' to BC. [Prop. I., p. 143.

13. Corresponding Points applied to circles of curvature. It has been shown that parallel straight lines correspond to parallel straight lines. [Prop. III., p. 144.

By a very similar process it may be proved that, if two straight lines PV, PT be equally inclined to the axis in opposite directions, the corresponding lines will be equally inclined to the axis in opposite directions.

Let the circle of curvature at P, in an ellipse, cut the ellipse in V, and let the tangent at P meet the axis in T. Take points p, v (on the auxiliary circle), corresponding to P, V. Then pv and the tangent pT are equally inclined

to the axis of the ellipse, since they correspond to lines which are equally inclined to the axis.

[Ex. 7, p. 158. Ex. 1. If the circles of curvature at the extremities P, D of two conjugate diameters of an ellipse meet the curve again in Q, R respectively, then PR is parallel to DQ.

This theorem may be reduced to the following:

If from the extremities p, d of two diameters at right angles, in a circle, the chords pq, dr be drawn equally inclined with the tangents at p, d respectively to a fixed diameter, then pr is parallel to dq.

Ex. 2. Any point 0 on a circle being given, it may be shown that there are three points P, Q, R on the circumference, situated at the vertices of an equilateral triangle, such that OP, OQ, OR are equally inclined with the tangents at P, Q, R to a given diameter.

It follows that there are three points P, Q, R on an ellipse, the circles of curvature at which pass through a given point on the curve, &c. [Ex. 13, p. 159. Def. If the opposite sides of any quadrilateral (ABCD) be produced to meet (in O, P) the figure thus formed is called a Complete Quadrilateral.

The straight lines AC, BD, OP are the Diagonals of the complete quadrilateral.

14. The middle points of the diagonals of a complete quadrilateral lie on the same straight line.

Complete the quadrilateral ABCD and let OP be the exterior diagonal.

Complete the parallelograms BODR, AOCQ, and let BR AQ cut PD in V, T respectively.

Then, since AQ is parallel to OC,

PC: CT=PB: BA

[Euc. VI., 2.

= PV : VD, similarly.

Hence, PQR is a straight line and the middle points of OP, OQ, OR lie upon a straight line parallel to PQR. [Euc. VI., 2.

But the middle point of OQ is also the middle point of AC, since the diagonals of parallelograms bisect one another. Similarly, the middle point of OR is also the middle point of BD.

Therefore the middle points of AC, BD, and OP lie on a straight line parallel to PQR.

N

CHAPTER XIII.

ANHARMONIC RATIO.

Let A, B, C, D be four points on a straight line. Then the ratio AB. CD: AD.BC is called the Anharmonic Ratio of the range A, B, C, D, and is denoted by {ABCD}.

In the expression for the anharmonic ratio of a range the letters which constitute the range might have been taken in a different order. Thus AD. BC: AB.CD might have been defined as the anharmonic ratio of the above range. It is of course necessary to retain throughout any investigation the particular order adopted at its commencement.

1. If four fixed straight lines which meet in O be cut by any transversal in the points A, B, C, D, then will {ABCD} be constant.

Draw the straight line aBb, parallel to OD, and meeting OA, OB in a, b.

Then

D

B

AB: AD=αB: DO,

by similar triangles ADO, ABa.