« ForrigeFortsett »
(Argument ad absurdum.) Theorem. On the same straight D line, and on the same side of it, there cannot be two similar segments of circles not coinciding with each other.
Steps of the Demonstration. Suppose that the segments ACB, ADB are similar, and on the same right line Al, and the same side of it, without coinciding with each other, and prove, on that supposition,
1. that one segment must fall within the other, 2. that _ ACB = _ ADB, i.e., ext. _ of A ACD =
int. 2, which is impossible, and :: the supposition is false.
PROPOSITION XXIV. Theorem. Similar segments of circles upon equal straight lines are equal to one another.
Steps of the Demonstration. Conceive the segments applied as directed, and Prove, 1. that AB coincides with cd, 2. that :: (by last Prop.) segment AEB = segPROPOSITION XXV. Problem. A segment of a circle being given, to describe the circle of which it is the segment.
ment CFD. * The remark, in Euclid, that if _ ABD > Bad the segment ABC would be < 0; and if _ ABD > BAD, the segment ABC would be > į O, is not necessary to the Demon stration, but may be considered as a corollary to it.
To prove Case 1st, in which _ ABD = _ BAD, show that DA, DB, DC = each other; and :. a o described with centre d and one of these lines as distance, will pass through the extremities of the other two.
Steps of the Demonstration to Case 2nd,
In which _ ABD +_ BAD. 1. Prove that AE = EB,
that (in AS ADE, CDE) base AE = base EC, that .. AE, EB, EC = each other; and :.
a o described with centre E and one of these lines as distance, will pass through the extremities of the other two*.
PROPOSITION XXVI. Theorem. In equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences.
Steps of the Demonstration. 1. Prove that (in As BGC, EHF) base bc = base EF,
• that segment Bac is similar to segment EDF,
that .. segt. BAC = segt. EDF, by xxiv. III. 4. that bkc = elf.
( Argument ad absurdum). Theorem. In equal circles, the angles which stand upon equal circumferences are equal to each other, whether they be at the centres or circumferences.
PROPOSITION XXVIII. Theorem. In equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less.
Steps of the Demonstration. 1. Prove that (in As BCK, EFL) _ BKC = _ ELF,
that :: BGC = EHF, by xxvi. III.
PROPOSITION XXIX. In equal circles, equal circumferences are subtended by equal straight lines.
Steps of the Demonstration. 1. Prove that BKC = _ ELF, . 2. that (in AS BKC, ELF) base bc=base EF.
PROPOSITION XXX. Problem. To bisect a given circumference, that is, to divide it into two equal parts.
....... ... 3 Steps of the Demonstration. 1. Prove that (in AS ACD, BCD,) base AD = base DB 2. that : AD = DB.
Theorem. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.