PROP. XXXIII. 43. PROBLEM. Upon a given finite straight line to describe a segment of a circle, which shall be similar to a given segment of another circle. Let ACB be a given segment of a circle, and B E DE a given finite straight line: It is required to describe on DE a segment of a circle, similar to the segment ACB. In ACB take any point C, and join A, C, and B, C: At the point D in DE make (E. 23. 1.) the ▲ EDF= < BAC; and, at the point E, also, make the 2 ABC; .. (S. 26. 1.) the About the A DFE describe < DFE DEF (S. 5. 1. cor.) the circle DFE; .. (E. 11. def. 3. and E. 21. 3.) the segment DFE is similar to the segment ACB. PROP. XXXIV, 44. THEOREM. If upon two opposite sides of an oblong, two similar segments of circles be described, the one of them lying wholly within the oblong, and the other wholly without it, the figure contained by the two remaining sides of the oblong and the two circular arches, shall be equal to the oblong. Upon the two opposite sides AD, BC, of the oblong ABCD, let there be described two similar segments of circles AED, BFC: the one, namely BFC, lying wholly within the oblong, and the other lying wholly without it: The figure contained by BA, AED, DC and CFB is equal to the oblong ABCD. For (hyp. and E. 34. 1.) AD=BC; ;. (hyp. and E. 24. 3.) the segment AED the segment BFC; to each of these equals add the figure ADCFB, and it is manifest that the figure AEDCFB is equal to the oblong ABCD. 45. COR. 1. An indefinite number of such mixtilineal figures may be found (S. 3. 1. cor. 3. and S. 33. 3.) equal to one another, and each of them equal to any given oblong. 46. COR. 2. If upon AB and DC, the two remaining sides of the oblong, there be, likewise, described two similar segments of circles ALB, DKC, it is evident that the figure ALBCDA is equal to the figure ABFCDEA; and that the figure ALBFCKDE=ABCD, ALB being supposed not to meet BFC again within ABCD. PROP. XXXV. 47. THEOREM. The arches of a circle that are intercepted between two parallel chords are equal to one another. Let AB and CG be two parallel chords of the circle ACGB: Then is AC = BG. L For join C, B: And, because (hyp.) CG is parallel to AB,... (E. 29. 1.) the GCB=2 CBA (E. 26. 3.) ACBG. PROP. XXXVI. 48. THEOREM. If two chords of a given circle intersect each other, the angle of their inclination is equal to the half of the angle at the centre standing upon the aggregate, or the difference, of the arches intercepted between them, accordingly as they meet within, or without the circle. First, let the two chords AB, CD, of the circle ACBD, cut one another in E, within the circle: The 4 DEB is equal to the half of an angle at the centre, standing upon a circumference equal to AC+DB. For through C draw (E. 31. 1.) CG parallel to AB; ... BG=AC, and DBG=AC+ DB; but (constr. and E. 29. 1.) the DEB DCG, which (E. 20.3.) is the half of an angle at the centre, standing upon DBG. Secondly, let the two chords DA and BC, meet when produced, without the circle, in F: If, then, AH be drawn parallel to CB, it may be shewn, in a similar manner, that the DFB is equal to the half of an at the centre standing on DH, which is the difference between AB and AC. PROP. XXXVII. 49. THEOREM. In equal circles the greater angle stands upon the greater circumference; whether the angles compared be at the centres or the circumferences. For whether the be at the centres, or the circumferences, if, from the greater, an ▲ (E. 23. 1.) be cut off equal to the less, the circumference on which it stands, will evidently be part of the circumference on which the greater stands, and will (E. 26. 3.) be equal to that on which the less ▲ stands; the which circumference is, .., less than the other. PROP. XXXVIII. 50. THEOREM. If from any given point, without a circle, there be drawn two straight lines cutting |