3. By means of Menelaus' Theorem prove that the following sets of three points are, in each case, collinear

(1) The points in which the three external bisectors of the angles of a triangle meet the opposite sides produced :

(2) The points in which two internal and one external bisector of the angles of a triangle meet the opposite sides (produced if necessary):

(3) Three points on the sides of a triangle (produced if necessary) if the three respectively equidistant from the mid points of the sides are collinear:

(4) The points in which two of the internal and one external bisector of the angles between three lines drawn from any point to the corners of a triangle meet its opposite sides.

4. By means of Desargues' Theorem prove that—

(1) When three triangles are in perspective, two and two, and have a common axis of perspective, their three centres of perspective are collinear:

(2) When three triangles are in perspective, two and two, and have a common centre of perspective, their three axes of perspective are concurrent. (Chasles, Géométrie Supérieure, pp. 283, 284: see also Townsend, Vol. I, p. 194, where these Theorems are generalised for any rectilineal figure.)

5. Prove the last Exercise, as in the Note to Theorem (4) by considering the triangles not in a plane.

6. Three circles touch the sides of a triangle ABC at the points where the in-circle touches them; and the circles touch each other in P, Q, R; show that AP, BQ, CR are concurrent.

7. Lines from two corners of a triangle divide the opposite sides in same ratio; if the third corner is joined to their intersection, this join produced will either bisect the third side, or divide it in duplicate of the above ratio.

8. ABC is a triangle, O a point within it; if AO, BO, CO meet BC, CA, AB in X, Y, Z respectively; and YZ, CB, produced meet in P; then

Z;

and

9. If a variable transversal cuts the sides of a fixed triangle in X, Y, the ratio of XY to YZ is fixed; then the circum-circle of any one of the three triangles cut off by the transversal goes through a fixed point.

10. If through one of the points of intersection of two circles two lines YXZ, QPR are drawn at right angles; so that X, P are on the line of centres; Y, Q on one circumference; and Z, R on the other; then

II. From the corners B, C, of a triangle ABC, BX, CY are drawn parallel to the opposite sides, so as each to be equal to a given length; if parallels to the adjacent sides, through X, Y, meet in Z; then XC, YB, ZA are concurrent.

12. If from any point within a triangle perpendiculars are dropped on its sides, and the circle through the feet of these perpendiculars cuts the sides again; then the perpendiculars at these last points of section are concurrent. 13. The perpendiculars from the corners of a triangle, on the sides of its pedal triangle, are concurrent at the circum-centre of the original triangle.

14. If twelve perpendiculars are drawn to the sides of a triangle, from the four centres of its circles of contact, these perpendiculars will be concurrent, three by three, in four points which are the centres of the circum-circles of the four ex-central triangles.

NOTE-The four ex-central triangles are formed by joining the ex- and in-centres.

15. In any triangle, show that the N. P.-circle bisects all lines from the orthocentre to the circumference of the circum-circle; and also bisects all lines from a corner to the circumference of the circle through the orthocentre and the other two corners: hence prove that

If in each ex-central triangle, its orthocentre is joined with its circum-centre, the four joins are concurrent in the centre of the N. P. circle of the four ex-central triangles.

NOTE-The last five Exercises are proved in Booth's New Geometrical Methods, Vol. II, pp. 263, 285, 300, 319, 320.

16. If A, B, C are three collinear points, and a, b, c three other collinear points; then the intersections of Ab, Ba, of Bc, Cb, and of Ca, Ac, are collinear.

NOTE-Use Theorem (8).

17. If A, B, C are three concurrent lines, and a, b, c three other concurrent lines; then (denoting the intersection of A and a by Aa) the joins of Ab, Ba, of Bc, Cb, and of Ca, Ac are concurrent. (Chasles, Géométrie Supérieure, p. 294.)

SECTION iii-CENTRES OF SIMILITUDE.

Note-By OA, it is to be understood that we mean to indicate the circle whose centre is A; where, from the construction of the figure, no ambiguity can arise.

Def. If the join of the centres A and B, of two circles, is divided externally in S, and internally in σ, so that

=

A: σ B,

SA: SB = radius of OA: radius of O B the point S is called the external centre of similitude, and the point σ is called the internal centre of similitude of the two circles.

THEOREM (1)—A tangent drawn to one of two circles, from either centre of similitude, is also a tangent to the other.

From S, the ext. centre of simil. of Os A, B, let SP be drawn, touching O A in P; and let BQ be

Then by simr. As

AP: BQ

=

on SP (produced if necessary).

SA: SB = radius of OA: radius of O B.

.. BQ = radius of O B.

.. SPQ is tang. to O B, at Q.

The proof is exactly simr. for σ, the int. centre of simil.

Def. The common tangents to two circles from their external centre of similitude are called their direct common tangents; and those from their internal centre of similitude are called their transverse common tangents.

Note-When one of the circles is entirely within the other, the circles have no common tangents, and therefore both centres of similitude lie within both circles.

Similarly when the circles intersect, the external centre of similitude lies outside both circles, and the internal centre of similitude within both.

And when the circles touch externally, the internal centre of similitude coincides with the point of contact.

THEOREM (2)—The join of the extremities of two parallel radii of two circles goes through their external centre of similitude, when the radii are on the same side of the line of centres; and through their internal centre of similitude, when the radii are on opposite sides of the line of centres.

Let AP, BQ be || radii of Os A, B ; and on same side of AB.

S

.. S is ext. centre of simil.

Simrly. if AP, BQ' are on opposite sides of AB; PQ' will go through σ, the int. centre of similitude.

Def. If through either centre of similitude of two circles, a line is drawn cutting the circles; then, of the four points of section, the one which is nearer the centre of similitude on one circle is said to correspond to the one which is nearer on the other circle; and this pair of nearer points are called corresponding points. So also the pair of remoter points are called corresponding points. But a nearer point of one circle, and a remoter point of the other circle, are called non-corresponding points.

THEOREM (3)-If through a centre of similitude of two circles, a line is drawn cutting the circles, the radii to a pair of corresponding points are parallel.

Let S be a centre of simil. of Os A, B ; and let SPQpq cut C A in P, Q, and O B in p, q; so that P, p and Q, q are the pairs of corresponding points. Then, since SA: SB AP: Bp;

...

AP and Bp are ||.

Simrly. AQ and Bq are ||.

Cor. The tangents at P, p, being L to AP, Bp are ; and so also are the tangs. at Q, q.

THEOREM (4)—If through a centre of similitude of two circles, a line is drawn cutting the circles; then the rectangle under the distances of one pair of non-corresponding points from that centre of similitude, is equal to the rectangle under the distances of the other pair of non-corresponding points from that centre; and each of these rectangles is constant.

And Sp.Sq

=

sq. on tang. from S to O B, which is const.

.. SP. Sq is const.

and .. so also is SQ. Sp.

Simrly. for the internal centre of similitude.

Def. Each of these constant rectangles is called a rectangle of anti

similitude.