straight AD or CD, or else we would have OCLCD and OALAD.

But in this last case the whole circle except points A and C would be within

ADC.

B

For

the center O would be within ADC, being then on the bisector of ADC since, ▲ADC being isosceles, ▲ AOD = A COD [3 sides =]. Hence the point B would be within ▲ ADC, which (by 144) is impossible. But it is just as impossible that AD or CD (besides A or C) should have a point D' on the circle other than D. For then we would have ADC=AD'C, which (by 79) is impossible.

FIG. 60.

146. Theorem. If two opposite angles of a quadrilateral are supplemental it is cyclic.

147. Corollary to 146. A quadrilateral is cyclic

if an angle is congruent to the angle adjacent to its opposite.

Ex. 107. Defining a tanchord angle as one between a tangent to a circle and a chord from the point of contact, prove it congruent to an inscribed angle on this chord.

Ex. 108. An angle made by two chords is how related to the angles at the center on chords joining the endpoints of the given chords?

Ex. 109. The vertices of all right-angled triangles on the same hypothenuse are concyclic.

Ex. 110. Tangents to a circle from the same external point are congruent, and make congruent angles with the straight through that point and the center.

Ex. 111. Two congruent coinitial chords are symmetric with respect to the coinitial diameter.

Ex. 112. If triangles on the same base and on the same side of it have the angles opposite it equal, the bisectors of these angles are copunctal.

Ex. 113. The end-points of two congruent chords of a circle are the vertices of a symmetrical trapezoid.

Ex. 114. The chord which joins the points of contact of parallel tangents to a circle is a diameter.

Ex. 115. A parallelogram inscribed in a circle must have diameters for diagonals.

Ex. 116. Of the vertices of a triangle and its orthocenter, each is the orthocenter of the other three.

Ex. 117. At every point on the circle can be taken one, and only one, tangent, namely, the perpendicular to the radius at the point.

Ex. 118. The perpendicular to a tangent from the center of the circle cuts it in the point of contact.

Ex. 119. The perpendicular to a tangent at the point of contact contains the center.

Ex. 120. The radius to the point of contact is perpendicular to the tangent.

Ex. 121. An inscribed ||g'm is a rectangle.

Ex. 122. The bisector of any

of an inscribed quad' intersects the bisector of the opposite exterior on the . Ex. 123. The with one of the sides of a ▲ as diameter bisects the base.

Ex. 124. The radius is = to the side of a regular inscribed hexagon.

Ex. 125. If the opposite sides of an inscribed quad' be produced to meet, the bisectors of the s so formed

are 1.

Ex. 126. The circles on 2 sides of a 4 as diameters intersect on the third side (in the foot of its altitude).

Ex. 127. The altitudes of a ▲ are the

pedal ▲ (the feet of the altitudes).

bisectors of its

Ex. 128. AB a diameter; AC any chord; CD tangent; BD1CD, meets AC on

B(BA).

Ex. 129. Find a point from which the three rays to three given points makes.

Ex. 130. The circumos of 3 As made by 3 points on the sides of a A, 2 with their vertex, are copunctal.

V. The Archimedes Assumption.*

V. Let A, be any point on a straight between any given points A and B; take then the points A2, A3, A1, . . ., such that A, lies between A and A2, furthermore A, between A, and A,, further A, between A2 and A, and so on, and also such that the sects AA1, A,A2, A2A3, A,A,, . . ., are congruent; then in the series of points A2, A, A,,..., there is always such a point An, that B lies between A and An.

148. This postulate makes possible the introduction of the continuity idea into geometry. We have not used it, and will not, since the whole of the ordinary school-geometry can be constructed with only Assumptions I–IV.

* Archimedis Opera, rec. Heiberg, vol. I, 1880, p. 11.

CHAPTER VI.

PROBLEMS OF CONSTRUCTION.

Existence theorems on the basis of assumptions I-V, and the visual representation of such theorems by graphic constructions.

Graphic solutions of the geometric problems by means of ruler and sect-carrier.

149. Convention. What are called problems of construction have a double import. Theoretically they are really theorems declaring that the existence of certain points, sects, straights, angles, circles, etc., follows logically by rigorous deduction from the existences postulated in our assumptions. Thus the possibility of solving such problems by elementary geometry is a matter absolutely essential in the logical sequence of our theorems.

So, for example, we have shown (in 101) that a sect has always trisection points, and this may be expressed by saying we have solved the problem to trisect a sect. Now it happens that a solution of the problem to trisect any angle is impossible with only our assumptions. Thus any reference to results following from the trisection of the angle would be equivalent to the introduction of additional assumptions.

But problems of construction, on the other hand, may have a reference to practical operations, usually for drawing on a plane a picture which shall serve as an approximate graphic representation of the data and results of the existential theorem.

Our Assumptions I postulate the existence of a straight as the result of the existence of two points. This may be taken as authorizing the graphic designation of given points and the graphic operation to join two designated points by a straight, and as guaranteeing that this operation can always be effected. Confining ourselves to plane geometry, on the basis of the same Assumptions I, we authorize the graphic operation to find the intersection-point of two coplanar non-parallel straights, and guarantee that this may always be accomplished.

To practically perform these graphic operations, that is for the actual drawing of pictures which shall represent straights with their intersections, we grant the use of a physical instrument whose edge is by hypothesis straight, namely, the straight-edge or ruler.

Thus Assumptions I give us as assumed constructions, or as solved, the fundamental problem of plane geometry:

Problem 1. (a) To designate a given point of the plane; (b) to draw the straight determined by two points; to find the intersection of two non-parallel straights.

150. Our Assumptions III postulate the existence on any given straight from any given point of it toward a given side, of a sect congruent to a given sect.