PROPOSITION XXXI. THEOREM. In a circle, the angle in a semicircle is a right angle; and 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. B Let ABC be a O, of which O is the centre and BC a diameter. Draw AC, dividing the into the segments ABC, ADC. Join BA, AD, DC. Then must the in the semicircle BAC be a rt. 4, and ▲ in segment ABC, greater than a semicircle, less than a rt. 4, and in segment ADC, less than a semicircle, greater than a rt. ▲ . First, the flat angle BOC=twice / BAC, III. C. p. 150. .. BAC is a rt. 4. Next, BAC is a rt. 4, .. ABC is less than a rt. 2. Lastly, sum of 4 s ABC, ADC=two rt. 2 s, I. 17. III. 22. Q. E. D. BOOK IV. INTRODUCTORY REMARKS. EUCLID gives in this Book of the Elements a series of Problems relating to cases in which circles may be described in or about triangles, squares, and regular polygons, and of the last-mentioned he treats of three only: the Pentagon, or figure of 5 sides, The Student will find it useful to remember the following Theorems, which are established and applied in the proofs of the Propositions in this Book. I. The bisectors of the angles of a triangle, square, or regular polygon meet in a point, which is the centre of the inscribed circle. II. The perpendiculars drawn from the middle points of the sides of a triangle, square, or regular polygon meet in a point, which is the centre of the circumscribed circle. III. In the case of a square, or regular polygon the inscribed and circumscribed circles have a common centre. IV. If the circumference of a circle be divided into any number of equal parts, the chords joining each pair of consecutive points form a regular figure inscribed in the circle, and the tangents drawn through the points form a regular figure described about the circle. 179 PROPOSITION I. PROBLEM. In a given circle to draw a chord equal to a given straight line, which is not greater than the diameter of the circle. B F D Let ABC be the given O, and D the given line, not greater than the diameter of the O. It is required to draw in the ABC a chord=D. Draw EC, a diameter of ABC. Then if EC=D, what was required is done. But if not, EC is greater than D. From EC cut off EF=D, and with centre E and radius EF describe a O AFB, cutting ABC in A and B ; and join AE. the Then, . E is the centre of AFB, .. EA=EF, and .. EA=D. Thus a chord EA equal to D has been drawn in © ABC. Q. E. F. Ex. Draw the diameter of a circle, which shall pass at a given distance from a given point. PROPOSITION II. PROBLEM. In a given circle to inscribe a triangle, equiangular to a given triangle. H D Let ABC be the given O, and DEF the given A. It is required to inscribe in ○ ABC a ▲, equiangular to ▲ DEF. Draw GAH touching the ABC at the pt. A. GAH is a tangent, and AB a chord of the O, .. 4 ACB= 4 GAB, that is, ▲ ACB= ▲ DFE. So also, ▲ ABC= 2 HAC, III. 17. I. 23. III. 32. III. 32. that is, ABC= ▲ DEF; .. remaining 4 BAC remaining 4 EDF; ..▲ ABC is equiangular to ▲ DEF, and it is inscribed in the ABC. Q. E. F. Ex. If an equilateral triangle be inscribed in a circle, prove that the radii, drawn to the angular points, bisect the angles of the triangle. PROPOSITION III. PROBLEM. About a given circle to describe a triangle, equiangular to a given triangle. B H Let ABC be the given C, and DEF the given ▲. It is required to describe about the a▲ equiangular to ▲ EDF. From O, the centre of the O, draw any Produce EF to the pts. G, H. radius OC. Make 4 COA ▲ DEG, and ▲ COB= ▲ DFH. = I. 23. Through A, B, C draw tangents to the O, meeting in L, M, N. Then will LMN be the ▲ required. For: ML, LN, NM are tangents to the O, Now 4s of quadrilateral AOCM together four rt. III. 18. ≤ s.; .. sum of 4 s COA, AMC=two rt. 4 s. and L Similarly, it may be shewn that ▲ LNM= ▲ DFE; .. also MLN LEDF. = I. 32. Thus a▲, equiangular to ▲ DEF, is described about the . Q. E. F. |