Every chord perpendicular to this axis is bisected by it.

If the circles have a common point on this straight they cannot have any other point in common, for any point in each has in that its symmetrical point with regard to this axis, and circles with three points in common are identical.

199. Two circles with one and only one point in common are called tangent, are said to touch, and the common point is called the point of tangency or contact.

200. If two circles touch, then, since there is only one common point, this point of contact is on the center-straight, and a perpendicular to the centerstraight through the point of contact is a common tangent to the two circles.

Ex. 175. Two circles cannot mutually bisect.

Ex. 176. The chord of half a minor arc is greater than half the chord of the arc.

Ex. 177. In a circle, two chords which are not both diameters do not mutually bisect each other.

Ex. 178. All points in a chord are within the circle. Ex. 179. Through a given point within a circle draw the smallest chord.

Ex. 180. Rays from center to intersection points of a tangent with || tangents are 1.

Ex. 181. A circle on one side of a triangle as diameter passes through the feet of two of its altitudes.

Ex. 182. In ▲ ABC if D on AB=BC, prove CD> AD. Ex. 183. A circumscribed parallelogram is a rhombus. Ex. 184. In ▲AABC, having AB> BC, the median BD makes BDA obtuse.

Ex. 185. If AB, a side of a regular ▲, be produced to D, then AD>CD> BC.

Ex. 186. If BD is bisector tь, and AB> BC, then BC> CD.

Ex. 187. How must a straight through one of the common points of two intersecting circles be drawn in order that the two circles may intercept congruent chords on it?

Ex. 188. Through one of the points of intersection of two circles draw the straight on which the two circles intercept the greatest sect.

Ex. 189. If any two straights be drawn through the point of contact of two circles, the chords joining their second intersections with each circle will be on parallels.

Ex. 190. To describe a O which shall pass through a given point, and touch a given in another given point.

Ex. 191. To describe a O which shall touch a given O, and touch a given st' [or another given O] at a given point.

Ex. 192. The foot of an altitude bisects a sect from orthocenter to circum-0.

Ex. 193. If from the end-points of any diameter of a given is be drawn to any secant their feet give with the center = sects.

Ex. 194. A, B, I, Ic are concylic.

Ex. 195. If to meets circum-0 in D, then DA =DC =DI. Ex. 196. The is at the extremities of any chord make = sects on any diameter.

Ex. 197. If in any 2 given tangent Os be taken any 2 || diameters, an extremity of each diameter, and the point of contact shall be costraight.

Ex. 198. If 2 Os touch internally, on any chord of one tangent to the other the point of contact makes sects which subtend = s at the point of tangency of the Os. Ex. 199. 2ma>

obtuse.

<a according as A acute, r't,

Ex. 200. Chords joining the end-points of || chords

are.

Ex. 201. St' through point of tangency meets O O at A, OO' at A'. Prove AO || A'O'.

Ex. 202. Intersecting Os are with regard to their center-st' and if are with regard to their common chord.

Ex. 203. Find the bi' of an without using its vertex. Ex. 204. A quad' with 2 sides || and the others = is either a g'm or a symtra.

CHAPTER VIII.

A SECT CALCULUS.

201. On the basis of assumptions I 1, 2, and II-IV, that is, in the plane and without the Archimedes assumption, we will establish a sect calculus or geometric algorithm for sects, where all the operations are identical with those for real numbers. The following proof is due to F. Schur.

202. (Pascal.) Let A, B, C and A', B', C' be two triplets of points situated respectively on two perpendiculars and distinct

from their intersection

point O'. If AB' is par- B

allel to A'B and BC' parallel to B'C, then is also AC' parallel to A'C.

Proof. Call D' the point where the perpendicular from B upon the straight A'C meets the straight B'A'. Then C is the orthocenter of the triangle BA'D'; therefore D'CLA'B and .. LAB'.

Consequently C is also the orthocenter of the tri

angle AB'D'; .. AD' 1 B'C and .. also 1 BC'. Consequently B is the orthocenter of triangle AC'D'; .. AC'L D'B and .. AC' || A'C.

203. Instead of the word " congruent" and the sign, we use, in this sect calculus, the word 'equal" and the sign =.

204. We begin by showing how from any two sects to find unequivocally a third by an operation we will call addition.

205. If A, B, C are three costraight points, and B lies between A and C, then we designate c = AC as the sum of the two sects a = AB and b=BC, and write to express this c = a+b.

To add the two sects a and b in a determined order, we start from any point A, and take the point B such that AB, that is = a. Then on the straight AB beyond B we take the point C such that BC=b. Then the sect AC is what we have designated as the sum of the two sects a = AB and b = BC in the order a+b.

206. From III 3 follows immediately that this sum is independent of the choice of the point A, and independent of the choice of the straight AB.

By III 1, it is independent of the order in which the sects are added. Therefore a+b=b+a.

207. This is the commutative law for addition. Thus the commutative law for addition holds good,