COR. From the hypothesis, that the exterior vertical angle, CAF is bisected by GA, which also cuts the base produced BCG in G and the circle in H, the conclusion follows, that the rectangle of the sides together with the square of the line which bisects the exterior angle is equal to the rectangle of the whole line produced and the extreme segment; i. e BA.AC + AG2 = BG.GC. SCH. Prop B and its Corollary combined, may be thus enunciated; “if the vertical or ext.vertical / of a ▲ be bisected by a st. line, which cuts the base, or the base produced, the square of the st. line shall be equal to the difference of the rectangles under the two sides and under the segments of the base or of the base produced." In the case of the bisection of the vert. And ext. vert. 66 BAC, AD2 = BA. AC-BD. DC. PROP. C.-THEOR. If from any angle of a triangle a st. line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle. CON. 5, IV. DEM. 31, III. 21, III. 4, VI. 16, VI. About ▲ ABC desc. O ACB and its diam. AE, & join EC. D. 131,III. & 21 III.·.· rt. ▲ BDA = ▲ ECA, & ABD 2 4, VI. 316, VI. = ▲ AEC; .. ▲ BDA is eq. ang. with ▲ ACE; & .. BA: AD = EA: AC; .. BA. AC - EA. AD. If therefore from any angle, &c. Q. E. D. COR.-If two ▲s, ABC, ACE, be inscribed in the same or in equals, the rect. under the two sides of the one, BA. AC, shall be to the rect. under the two sides of the other, EC. CA, as the perp. AD, which is drawn from the vertex A, to the base, BC, of the one, is to the perp.which is drawn from the vertex C to the base, AE, of the other i. e. BA. AC: EC. CA =AD: perp. from C. PROP. D.-THEOR. The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles contained by its opposite sides. CON. 23, I. DEM. 21, III. 4, VI. 16, VI. 1, II. If there be two st. lines, one of which is divided into any number of pts.; the rect. contained by the two st. lines is equal to the rectangles contained by the undiv. line and the several pts. of the divided line. .. ZABD = EBC. 221, III. and ·.· / BDA = / BCE .. ▲ ABD is eq. ang. with ▲ EBC; and .. BC. DA — BD. CE. Again, · ▲ ABE = ▲ DBC, and ▲ BAE = < BDC, ..A ABE is eq. ang with ▲ BCD, 34, VI. . BC: CE 4 C.21, III 5 4, VI. 6 16, VI. 7 D. 3. 81, II. 9 Rec. .. the whole .. BA: AE = BD: DC; AC.BD = AB. CD + AD.BC. Therefore, the rectangle contained, &c. Q. E. D. SCH. This Proposition is named PTOLEMY's Theorem, and is a Lemma at p. 9 of his Almagest, or Mɛyzλn Euvragıs of which it is said, “Ptolemy appears as a splendid mathematician, and an (at least) indifferent observer." 66 It is curious to note how editor after editor of Euclid has followed the identical diagrams of the earliest printed editions. PTOLEMY'S thirteen books Magne Constructionis, Id est, Perfectæ cælestium motuum, pertractationis," published at Basle by Symon Grynoeus in the year 1538, contain several diagrams, and the one to Prop. D, bk. VI, among the number, which have scarcely undergone any change since that time. Billingsley's English Euclid also furnishes very many examples from which to prove the imitativeness of succeeding Editors, and by which to justify, if need be, the continuation of the practice. PROP. E.-THEOR. The diagonals, AB, CD, of a quadrilateral ACBD, inscribed in a circle, ABC are to one another as the sums of the rectangles under the sides adjacent to the extremities of those diagonals: i. e., AB: CD: CA. AD + CB. BD: AC. CB + AD. DB. CON. 12, I. DEM. Cor. C, VI. Sch. 3, 24, V. In any number of magnitudes of the same kind forming two series, if the ratios of the 1st to the 2nd, of the 2nd to the 3rd, of the 3rd to the 4th, and so on, be the same in the two series; then any two combinations whatever, shall be to one another as two similar combinations of the corresponding magnitudes of the second series. 4, 5, VI. 11, V. C. Sup. Let AB, CD, cut one another in the . E. CASE I. Suppose that AB cuts CD at rt. Ls. CASE II. Or, Suppose that AB cuts CD not at rt. Ls. C. 12, I. Draw the perps. Aa, Bb, Cc, Dd. D. 1 Sim. 2 C. As before, Aa + Bb: Cc + Dd = AC.AD + but As AEa, BEb, CEc, and DEd are eq. ang.; D. 3 4, 5, VI. .. Aa, Bb, Cc, Dd, are to one another, as AE, BE, CE and DE; .. Aa, + Bb: Cc, + Dd = AE + BE: CE + DE =AB: CD. 4 6 Rec. = + DA. DB. AC. AD + BC. BD: CA. CB Therefore, the diagonals of a quadrilateral, &c. Q. E. D. USE AND APP. By aid of the last two Propositions, D & E, bk. VI, the following Problem may be solved. Given four st. lines, any three of which are together greater than the fourth, to construct a quadrilateral, of which the sides shall be equal to those four given st. lines, in a given order, each to each, and of which also its angular points lie in the circumference of a circle. By E, VI. The ratio of the diagonals is given; they are as the sums of the rectangles under the sides; By D, VI. The rectangle of the diagonals is given, being equal to the sum of the rectangles under the opp. sides; By Prob. 5 Use and App. 13, VI., we find the two st. lines; in this case, they are the two diagonals equal to a given rectangle ; And by 22, I, having now the sides and diagonals of the quadrilateral, we construct two triangles the sum of which will equal the required quadrilateral. PROP. F.-THEOR. If AB, a segment of a circle, ABD, be bisected in C, and from the extremities, A, B, of the base of the segment, and from C the point of bisection, st. lines be dawn to any point D in the circumference; the sum of the two lines AD + DB drawn from the extremities of the base, will have to the line DC drawn from the point of bisection the same ratio which the base BA, of the segment has to AC the base of half the segment. |