IN PROPOSITION XXVI. THEOREM XXIII. N equal circles (BADM, EFGN,), equal angles, whether they be at the centers as (C & H) or at the circumferences as (A & F), ftand upon equal arches (BMD, ENG,). T DEMONSTRATION. HE two fides CB, CD, of the ▲ BCD being to the two fides HE, HG, of the ▲ EHG (Hyp. 3. & Ax. 1. B. 3.), & the VC= to the VH (Hyp. 2.). 2. The fegment BAD is fimilar to the fegment EFG. 3. Wherefore, the base BD being fegments will be to one another. to the bafe EG (Arg. 1.), these Therefore, if the equal fegments BAD, EFG, (Arg. 3.) be taken away from the equal BADM, EFGN, (Hyp. 3.). P. 4. B. 1. Ax.2. B.3. P. 24. B. 3. 4. The remaining arches BMD, ENG, will be alfo to one another. Ax. 3. B. 1. Which was to be demonftrated. PROPOSITION XXVII. THEOREM XXIV. IN equal circles (BAG, DEF,) the angles, whether at the centers as (BCG & H) or at the circumferences as (A & E), which stand upon equal arches (BG, DF,); are equal to one another. If not, DEMONSTRATION. The V BCG & H at the centers will be unequal, & one, as Preparation. At the point Є in the line BC, make the V BCK to V H. THEREFORE the arch BK is to the arch DF. But the arch DF being to the arch BG (Hyp. 2) 2. The arch BK will be alfo = to the arch BG; which is impoffible { 3. Confequently, the V BCG & H at the centers, are to one another. Which was to be demonftrated. I. And these being double of the VA & F at the O (P. 20. B. 3.). 4. Thefe VA & E at the O, are alfo to one another. P. 23. B. 1. P. 26. B.3. Ax. 1. B.1. Ax.7. B.1. IN PROPOSITION XXVIII. N equal circles (ABDE, FHMN,); the equal chords (AD, FM,) fubtend equal arches (ABD, FHM or AED, FNM,). Hypothefis. 1. The O ABDE, FHMN, are equal. 11. The chords AD, FM, are equal. Thefis. The chords AD, FM, fubtend equal arches ABD, FHM or AED, FNM. THEOREM XXV. Preparation. 1. Find the centers C & G of the two ABDE, FHMN. DEMONSTRATION. BECAUSE the O ABDE, FHMN, are equal (Hyp. 1.). 1. The fides CA, CD, & GF, GM, of the ▲ ACD, FGM, are equal. And the chords AD, FM, being equal (Hyp. 2.). 2. The V ACD, FGM, are = to one another. 3. Confequently, the arches AED, FNM, fubtended by the chords AD, FM, will be alfo to one another. P. 1. B. 3. Ax. 1. B.3. P. 8. B. 1. P. 26. B. 3. Ax. 3. B. 1. Which was to be demonftrated. 4. And moreover, the whole O being equal (Hyp. 1.), the arches ABD, FHM, are alfo equal. PROPOSITION XXIX. THEOREM XXVI. N equal circles (BADM, EFHN,); equal arches (BMD, ENH,) are fubtended by equal chords (BD, EH,). BECAUSE DEMONSTRATION. ECAUSE the O BADM, EFHN, are equal (Hyp. 1.). 1. The fides CB, CD, & GE, GH, of the ▲ BCD, EGH, are to one another. But the arches BMD, ENH, being alfo equal (Hyp. 2.). P. 1. B. 3. Ax. 1. B. 3. P. 27. B. 3. 3. Confequently, the chord BD is to the chord EH. P. 4. B. 1. Which was to be demonftrated. 2. The VC & G, contained by thofe equal fides, will be to one another. To PROPOSITION XXX. PROBLEM IV. O divide an arch (ABD) into two equal parts (AB, BD,). Given The arch ABD. Refolution. Sought The divifion of the arch ABD into two equal parts AB, BD. 1. From the point A to the point D, draw the chord AD. Pof. I. P. 10. B. 1. 3. At the point C in the straight line AD, erect the CB, which P. 11. B. 1. Preparation. Draw the chords AB, DB. BECAUSE ECAUSE the fide AC is Pof. 1. to the fide CD (Ref. 2), CB com mon to the two ▲ ABC, DBC, & the V ACB to the V DCB (Ax. 10. B. 1. & Ref. 3.). 1. The bafe AB is to the bafe DB. P. 4. B. 1. 2. Confequently, the arches AB & DB, fubtended by the equal chords AB, DB, are to one onother, and the whole arch ABD, is divided into two equal parts in B. P. 28. B. 3. Which was to be done. |