circles a diameter of each circle is drawn.

joining the other ends of the diameters pass common point.

Shew that the lines

through the other

*62. P is any point on a circular arc APB. Shew that the bisector of the angle APB passes through a fixed point, and that the bisector of its supplementary angle passes through another fixed point.

63. The perpendiculars from A and C to the opposite sides of a triangle ABC intersect in E, and BD is the diameter through B of the circumscribing circle. Prove that AE is equal to CD, and that AC,ED bisect each other.

64. Shew that the circumscribing circles of the equilateral triangles described on the sides of any triangle and external to the triangle meet in a point.

65. Upon the sides of the triangle ABC the equilateral triangles A'BC, B'CA, and C'AB are constructed, the vertices A and A' being on opposite sides of BC, and so of the rest. Prove that the lines AA', BB', CC' meet in a point.

66. Prove that the points determined in Exercises 64 and 65 are the same, and if that point be D, that AA' = BB' = DA + DB + DC.

= CC'

Any

67. A circle is described on OA as diameter. line through O meets the circumference in P and the perpendicular to OA through A in Q. Shew, by Book II., Ex. 19, that the rectangle OP, OQ is constant. Also enunciate and prove the converses.

68. Squares are described on two sides of a triangle as bases, and on the third side as diagonal. Shew that the three circles about these squares have a point in common.

69. If a hexagon ABCDEF be inscribed in a circie

the sum of the angles A, C, E is equal to that of the angles B, D, F.

Enunciate and prove a similar theorem for any rectilineal figure with an even number of sides inscribed in a circie.

70. Two parallel chords AB and CD are drawn in a circle ABCD, whose centre is O. The chord CD is bisected in F, and a circle is described through A, O, F, cutting the given circle again in G. Prove that B, F, G lie in a straight line.

7J. Of all triangles which have a given base and vertical angle, the greatest is that in which the two sides are equal.

72. Two circles intersect at the points A and B. In the circumference of one of the circles ABC any point P is taken, and the straight lines PA, PB (produced when necessary) meet the circumference of the other circle at the points Q, R. Shew that the chord QR will be of the same length whatever may be the position of P.

73. Through one of the points of intersection of two equal circles, each of which passes through the centre of the other, a line is drawn to intersect the circles in two other points. Prove that these points and the other point of intersection of the circles are the angular points of an equilateral triangle.

74. Through each of the points of intersection of two circles straight lines are drawn, the one meeting the circles in points A and B, and the other meeting them in points C and D. Prove that AC is parallel to BD.

75. If two triangles have their bases, areas, and angles at the vertices equal, they are equal in all respects.

SECTION V.

TANGENTS.

THEOR. 20. Of all straight lines passing through a point on the circumference of a circle there is one, and only one, that does not meet the circumference again, and this straight line is perpendicular to the radius to the point.

Let A be a point on the circumference of the circle ABC whose centre is O, DAE the perpendicular through A to the radius OA :

B

then shall A be the only point in which DE meets the circumference.

Take any point F on DE, and join OF.

Because OA is perpendicular to DE,

therefore Of is greater than OA;

1. 15.

therefore F is without the circle ABC;

III. 1, Cor. therefore the straight line DF meets the circumference of the circle ABC in the point A only.

Again, let GAH be a straight line through A not perpendicu

lar to OA:

B

then shall GH meet the circumference in a second point:

Draw OK perpendicular to GH;

I. Prob. 3.

from O draw OL to meet GH at L, and making the angle KOL equal to the angle KOA.

1. Prol. 5.

Then, because OL and OA are equally inclined to OK the perpendicular to GH,

therefore OL is equal to OA;

that is, OL is equal to the radius of the circle;

therefore L is on the circumference ;

I. 1.

III. 1, Cor.

that is, the straight line GH meets the circumference in a second

point L.

Q.E.D.

DEF. 13. The straight line which, meeting the circumference of a circle in one point, does not meet it again, is said to touch, or to be a tangent to the circle at the point. The point is called the point of contact.

Let A be a point on the circumference of the circle ABC whose centre is O, GAH a straight line through A not perpendicular to OA:

then shall GH meet the circumference in a second point.

Draw OK perpendicular to GH, from O draw OL to mee GH at L and making the angle KOL equal to the angle KOA. I. Probs. 3 and 5.

Then, because OL and OA are equally inclined to OK the perpendicular to GH,

therefore OL is equal to OA;

that is, OL is equal to the radius of the circle;

I. 15.