SCHOLIUM. If the arcs are in equal circles, the proof is similar to that above; and, in general, the demonstration of a theorem concerning arcs, etc., in equal circles would be similar to that for arcs in the same circle, and vice versa.

PROPOSITION IX. THEOREM.

182. In the same circle, or in equal circles, equal chords are equally distant from the center; and conversely.

1o. Given: In circle ADB, chord AB equal to chord CD; To Prove: AB and CD are equally distant from O, the center.

Draw OP to AB, OR to CD, and join OA, OC.

Since OP is to AB, and OR is 1 to CD, (Const.)
AP = AB, and CR = 1⁄2 CD,

.. AP=CR (Ax. 7), and 04 = OC,

... rt. ▲ OAP = rt. ▲ OCR,

.. OPOR.

(172)

(162)

(72)

2°. Given: In circle ADB, chords AB, CD, equally distant from

... 2 AP or AB = 2 CR or CD. Q.E.D. (Ax. 6)

PROPOSITION X. THEOREM.

183. In the same circle, or in equal circles, a greater chord is nearer the center; and conversely.

B

1o. Given: In circle ADB, OP perpendicular to chord AB,

OQ perpendicular to chord CD, and AB greater than CD;

Since ABCD (Hyp.), arc AEB > arc CD.

On arc AEB lay off arc AE = arc CD.

Join AE, and draw OR to AE;

then AE CD (174'), and OR = OQ.

(181")

(178)

(92)

(182')

(Const.)

(Ax. 8)

(93)

Since AB lies between 0 and AE,

OR must cut AB in some point s,

.. OSOR.

But OPOS,

.. OP OR or OQ.

2°. Given: In circle ADB, OP perpendicular to chord AB, OQ perpendicular to chord CD, and OP less than 0Q;

To Prove: Chord AB is greater than chord CD.

Chord AB cannot be equal to chord CD, for then OP would be equal to 09 (182'); nor can AB be less than CD, for then OP would be greater than oQ (1°). Hence, since AB can be neither equal to nor less than CD, it must be greater than CD.

184. COR. A diameter is greater than any chord not passing through the center.

PROPOSITION XI. THEOREM.

185. Through any three given points not in the same straight line one circumference can be described, and only one.

D

B

Given: Three points A, B, C, not in the same straight line; To Prove: One circumference can be described through A, B, and C.

Join AB, BC, and at D, E, the mid points of AB, BC, respectively, draw DO 1 to AB, and EO to BC.*

(80) ·.· DO, EO, are to AB, BC, resp., at their mid points, (Const.) DO, EO, will meet at a point o equidistant from A, B, and C;

(156)

.. the circumference described from o as center, with radius OA, will pass through A, B, and C.

Q.E.D. (162)

Moreover, there can be but one such circumference.

For

the center of any circumference passing through A, B, and C', must lie both in DO and in EO (97), which lines can have but one common point.

186. COR. Two circumferences can intersect in not more than two points.

For if they could intersect in three points, there would be two different circumferences passing through the same three points, which is impossible.

187. SCHOLIUM. One circumference, and only one, can be described through the vertices of a given triangle.

* The construction by which we find the mid point of a line, evidently also gives the perpendicular at that point.

TANGENTS AND SECANTS.

188. A tangent is a straight line that touches a circumference in one point only; as ABC. The point B in which the tangent touches the circle is called the point of contact.

189. A secant is a straight line that cuts a circumference in two points; as DE.

190. A straight line perpendicular to a radius at its extremity is a tangent to the circle.

Given: AB perpendicular to OC, a radius of circle CEF, at its extremity C;

To Prove:

AB is a tangent to circle CEF.

Join o with D, any point in AB except C. Then

.. AB is tangent to CEF at C, Q.E.D. (188)

(since C is the only point in AB not without CEF.)

191. COR. 1. A tangent is perpendicular to the radius drawn to the point of contact.

For oc must be less than any other line drawn from o to AB (162).

192. COR. 2. A perpendicular to a tangent at the point of contact passes through the center of the circle.

For otherwise there could be two perpendiculars to the tangent at that point (191).

193. COR. 3. A perpendicular from the center to a tangent meets it at the point of contact (192).

194. COR. 4. At a given point of contact there can be but one tangent.

195. Parallels intercept equal arcs on a circumference.

Given Two parallels AB, CD, cutting or touching the circumference whose center is 0;

To Prove: These parallels intercept equal arcs.

1o. Let AB, CD, be secants.

Through 0, the center, draw the diameter MN to AB. Since AB is to CD (Hyp.), and MN is to AB, (Const.)

MN is to CD,

.. arc MA are MB, and arc MC arc MD,

(107) (175)

(Ax. 3)

(191)

(107)

Q.E.D.

(175)

2o. Let AB be tangent at M, and CD a secant.

Draw OM. Then OM is to AB

and also to its parallel CD;

.. arc MC arc MD.