PROPOSITION XXIII. (Argument ad absurdum.) Theorem. On the same straight line, and on the same side of it, there cannot be two similar segments of circles not coinciding with each other. D A B Steps of the Demonstration. Suppose that the segments ACB, ADB are similar, and on the same right line AB, 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 ▲ ACD = int., which is impossible, and .. the supposition is false. PROPOSITION XXIV. Theorem. Similar segments of circles upon equal straight lines are equal to one another. E 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 = seg ment CFD. PROPOSITION XXV. Problem. A segment of a circle being given, to describe the circle of which it is the segment. Το prove Case 1st, in which D ABD = BAD, show described that DA, DB, DC each other; and .. a 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, 1. Prove that AE = EB, 2. 3. that (in AS ADE, CDE) base AE = base EC, that .. AE, EB, EC each other; and .. a described with centre E and one of these lines as distance, will pass through the extremities of the other two*. The remark, in Euclid, that if ment ABC would be <; and if ABD > BAD the seg- BAD, the seg ment ABC would be >, is not necessary to the Demon stration, but may be considered as a corollary to it. PROPOSITION XXVI. Theorem. In equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences. B G D Steps of the Demonstration. 1. Prove that (in AS BGC, EHF) base BC = base EF, 2. 3. 4. ⚫ that segment BAC is similar to segment EDF, that.. segt. BAC = segt. EDF, by xxiv. III. that BKC ELF. PROPOSITION XXVII. (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. Steps of the Demonstration. State that if ▲ BGC = ≤ EHF, the BAC must = EDF, by xx. III. Then suppose that BGC EHF; and that BGC is the greater; and that EGK = EHF: and prove, on that supposition, 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, 2. that. BGC = EHF, by xxvi. III. 3. that BAC EDF. 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 вc=base EF. PROPOSITION XXX. D Problem. To bisect a given cir cumference, that is, to divide it into two equal parts. A Steps of the Demonstration. B 1. Prove that (in AS ACD, BCD,) base AD = base DB that. AD = DB. 2. |